Re: Help me an idiot
On Mon, 30 Apr 2001 20:25:36 -0700, James H. Steiger not@my_address wrote: If you read my post carefully, (assuming it made your server), you will see that it, indeed, gives the combinations solution you gave below, and shows that it is equivalent to another, more elegant way of arriving at the same answer, i.e., 2^N = (in this case) 2^5 = 32 It's amazing how much effort y'all put into doing some lazy pricks homework for him Gary Carson http://www.garycarson.com = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Re: Help me an idiot
I guess I shouldn't be surprised that comments in this thread can range from the sublime to the utterly tasteless from supposed professionals. Way to go. reg - Original Message - From: Gary Carson [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Tuesday, May 01, 2001 7:32 AM Subject: Re: Help me an idiot On Mon, 30 Apr 2001 20:25:36 -0700, James H. Steiger not@my_address wrote: If you read my post carefully, (assuming it made your server), you will see that it, indeed, gives the combinations solution you gave below, and shows that it is equivalent to another, more elegant way of arriving at the same answer, i.e., 2^N = (in this case) 2^5 = 32 It's amazing how much effort y'all put into doing some lazy pricks homework for him Gary Carson http://www.garycarson.com = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ = = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Re: Help me an idiot
Several have written to this thread and I believe there has been some misleading information passed along and intermixed with correct information. On Sun, 29 Apr 2001 11:26:11 -0400 Zina Taran [EMAIL PROTECTED] writes: I believe, the thrust of the fries reply was the overcount in the 5*4*3*. response rather than an expression of culinary preferences. Multiplying 5*4*3... is the kind of multiplication used when determining the 'number of permutations'. I seem to remember the question asked for number of combinations. The first step to obtaining the answer is to be sure you understand what you are 'counting.' If PERMUTATIONS - If you consider a hamburger with lettuce and mayo to be different than a hamburger with mayo and lettuce, then you want to count the number of permutations. Permutations involve the concept of order [rather it be 'the order in which you listed the condiments when placing the order' or 'the order in which the condiments are placed on your burger'] as an integral part of the problem. As applied to this particular problem, there were 5 condiments - for the first condiment selected there are 5 choices, for the second condiment selected there would be 4 choices available, and so on. Then by multiplying, you find the number of possibilities. I seriously doubt that you are interested in counting the number of permutations. If COMBINATIONS - If what you want to know is how many different combinations of condiments can be ordered, with no regards to an order concept, then you are interested in counting the number of combinations. It is this concept that I would think answers the question asked. One way to approach this is to look at and determine the count for each case, then find the total. [This approach was suggested in an earlier e-mail.] Five (5) condiments to choose from - 1) find the number of ways you can select exactly zero (0) condiments - there is 1, 2) find the number of ways you can select exactly one (1) condiment - there are 5, 3) find the number of ways you can select exactly two (2) condiments - there are 10, 4) find the number of ways you can select exactly three (3) condiments - there are 10, 5) find the number of ways you can select exactly four (4) condiments - there are 5, 6) find the number of ways you can select exactly five (5) condiments - there are 1, I'll leave it to you to find each of them. Answer: 1+5+10+10+5+1 = 32 different possible combinations of condiments can be ordered. If you are not interested in all the cases individually, there is a shorter way. Think of each different condiment that you can order, say pickles - you have 2 choices - 'order it' or 'don't order it'. The same two (2) options are available for each of the other condiments - catsup, mayo, lettuce, etc. [even for the fries and onion rings, if you want them included as a condiment on your burger. :-} ] Now what you have is a special case of the basic Multiplication Rule. Two (2) choices for each condiment, 2*2*2* ... *2, using 'n' 2's, or 2**n. Thus, if you have 5 condiments to choose from, there are 2**5 = 32 different combinations of condiments can be ordered for your hamburger. I hope this helps. - Robert R. Barbara S. Johnson E-mail: [EMAIL PROTECTED] Post: 84 West Lake Rd., Branchport, NY 14418 = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Re: Help me an idiot
On 30 Apr 2001 12:18:55 -0700, [EMAIL PROTECTED] (Robert R Johnson) wrote: Several have written to this thread and I believe there has been some misleading information passed along and intermixed with correct information. Possibly, you missed it, but I posted the correct answer last Saturday night telling the kid wanting help with his homework that it was somewhere near 32 combinations after another poster had inadvertently given him the no. of permutations. The only reason I said somewhere was so I would not bring down the wrath of those on the newsgroup opposed to giving a straight answer to a homework assignment :-))) Your response was quite helpful and a good learning tool. On Sun, 29 Apr 2001 11:26:11 -0400 Zina Taran [EMAIL PROTECTED] writes: I believe, the thrust of the fries reply was the overcount in the 5*4*3*. response rather than an expression of culinary preferences. Multiplying 5*4*3... is the kind of multiplication used when determining the 'number of permutations'. I seem to remember the question asked for number of combinations. The first step to obtaining the answer is to be sure you understand what you are 'counting.' If PERMUTATIONS - If you consider a hamburger with lettuce and mayo to be different than a hamburger with mayo and lettuce, then you want to count the number of permutations. Permutations involve the concept of order [rather it be 'the order in which you listed the condiments when placing the order' or 'the order in which the condiments are placed on your burger'] as an integral part of the problem. As applied to this particular problem, there were 5 condiments - for the first condiment selected there are 5 choices, for the second condiment selected there would be 4 choices available, and so on. Then by multiplying, you find the number of possibilities. I seriously doubt that you are interested in counting the number of permutations. If COMBINATIONS - If what you want to know is how many different combinations of condiments can be ordered, with no regards to an order concept, then you are interested in counting the number of combinations. It is this concept that I would think answers the question asked. One way to approach this is to look at and determine the count for each case, then find the total. [This approach was suggested in an earlier e-mail.] Five (5) condiments to choose from - 1) find the number of ways you can select exactly zero (0) condiments - there is 1, 2) find the number of ways you can select exactly one (1) condiment - there are 5, 3) find the number of ways you can select exactly two (2) condiments - there are 10, 4) find the number of ways you can select exactly three (3) condiments - there are 10, 5) find the number of ways you can select exactly four (4) condiments - there are 5, 6) find the number of ways you can select exactly five (5) condiments - there are 1, I'll leave it to you to find each of them. Answer: 1+5+10+10+5+1 = 32 different possible combinations of condiments can be ordered. If you are not interested in all the cases individually, there is a shorter way. Think of each different condiment that you can order, say pickles - you have 2 choices - 'order it' or 'don't order it'. The same two (2) options are available for each of the other condiments - catsup, mayo, lettuce, etc. [even for the fries and onion rings, if you want them included as a condiment on your burger. :-} ] Now what you have is a special case of the basic Multiplication Rule. Two (2) choices for each condiment, 2*2*2* ... *2, using 'n' 2's, or 2**n. Thus, if you have 5 condiments to choose from, there are 2**5 = 32 different combinations of condiments can be ordered for your hamburger. I hope this helps. - Robert R. Barbara S. Johnson E-mail: [EMAIL PROTECTED] Post: 84 West Lake Rd., Branchport, NY 14418 = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ = = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Re: Help me an idiot
If you read my post carefully, (assuming it made your server), you will see that it, indeed, gives the combinations solution you gave below, and shows that it is equivalent to another, more elegant way of arriving at the same answer, i.e., 2^N = (in this case) 2^5 = 32 Jim Steiger On 30 Apr 2001 12:18:55 -0700, [EMAIL PROTECTED] (Robert R Johnson) wrote: Several have written to this thread and I believe there has been some misleading information passed along and intermixed with correct information. On Sun, 29 Apr 2001 11:26:11 -0400 Zina Taran [EMAIL PROTECTED] writes: I believe, the thrust of the fries reply was the overcount in the 5*4*3*. response rather than an expression of culinary preferences. Multiplying 5*4*3... is the kind of multiplication used when determining the 'number of permutations'. I seem to remember the question asked for number of combinations. The first step to obtaining the answer is to be sure you understand what you are 'counting.' If PERMUTATIONS - If you consider a hamburger with lettuce and mayo to be different than a hamburger with mayo and lettuce, then you want to count the number of permutations. Permutations involve the concept of order [rather it be 'the order in which you listed the condiments when placing the order' or 'the order in which the condiments are placed on your burger'] as an integral part of the problem. As applied to this particular problem, there were 5 condiments - for the first condiment selected there are 5 choices, for the second condiment selected there would be 4 choices available, and so on. Then by multiplying, you find the number of possibilities. I seriously doubt that you are interested in counting the number of permutations. If COMBINATIONS - If what you want to know is how many different combinations of condiments can be ordered, with no regards to an order concept, then you are interested in counting the number of combinations. It is this concept that I would think answers the question asked. One way to approach this is to look at and determine the count for each case, then find the total. [This approach was suggested in an earlier e-mail.] Five (5) condiments to choose from - 1) find the number of ways you can select exactly zero (0) condiments - there is 1, 2) find the number of ways you can select exactly one (1) condiment - there are 5, 3) find the number of ways you can select exactly two (2) condiments - there are 10, 4) find the number of ways you can select exactly three (3) condiments - there are 10, 5) find the number of ways you can select exactly four (4) condiments - there are 5, 6) find the number of ways you can select exactly five (5) condiments - there are 1, I'll leave it to you to find each of them. Answer: 1+5+10+10+5+1 = 32 different possible combinations of condiments can be ordered. If you are not interested in all the cases individually, there is a shorter way. Think of each different condiment that you can order, say pickles - you have 2 choices - 'order it' or 'don't order it'. The same two (2) options are available for each of the other condiments - catsup, mayo, lettuce, etc. [even for the fries and onion rings, if you want them included as a condiment on your burger. :-} ] Now what you have is a special case of the basic Multiplication Rule. Two (2) choices for each condiment, 2*2*2* ... *2, using 'n' 2's, or 2**n. Thus, if you have 5 condiments to choose from, there are 2**5 = 32 different combinations of condiments can be ordered for your hamburger. I hope this helps. - Robert R. Barbara S. Johnson E-mail: [EMAIL PROTECTED] Post: 84 West Lake Rd., Branchport, NY 14418 = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ = = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Re: Help me an idiot
On Sat, 28 Apr 2001 20:48:03 +0800, Abdul Rahman [EMAIL PROTECTED] wrote: Please help me with my statistics. Question: If you order a burger from McDonald's you have a choice of the following condiments:ketchup, mustard , lettuce. pickles, and mayonnaise. A customer can ask for all thesecondiments or any subset of them when he or she orders a burger. How many different combinations of condiments can be ordered? No condiment at all conts as one combination. Your help is badly needed Just an Idiot@leftover This problem can be solved in two rather different, but equivalent ways. The fact that they ARE equivalent reveals a well-known, but surprising identity in combinatorics. One approach is as follows: To make the problem simpler, call the condiments A,B,C,D,E Ask, how many distinctly different ways you can construct burgers with NO condiments. [There is exactly one such way.] In combinatorial theory, you would call this 5 choose 0 the number of combinations of 5 objects taken 0 at a time. Next, ask, how many ways you can construct a burger with ONE condiment. [There are 5 such ways] In combinatorial theory, you would call this 5 choose 1. You proceed onward, enumerating all such ways. It is, of course, the sum, from i=0 to 5, of 5 choose i. This is the brute force method, and rather unrewarding and tedious. The other method is much simpler, and involves a straightforward insight. That is, any combination of condiments may be CODED with a binary digit code. Specifically, if a condiment is ON the burger, code it 1. If it is not on the burger, code it zero. So the burger with NO condiments would have the code 0 Some other examples A B C D E --- 0 0 000 (No condiments) 1 0 000 (Only condiment A) 0 1 000(Only condiment B) 1 1 000(Condiments A and B) 1 1 111(All 5 condiments) So you see, the answer to your problem can also be thought of in another way. That is, HOW MANY distinct 5 digit binary codes can you construct? Each unique code corresponds to a condiment selection. The fact that these two distinctly different approaches both yield correct answers proves a classic result, i.e. N Sum (N choose i) = 2^N i=0 Study what I have written carefully, and you should be able to arrive easily at the solution to your specific problem, and also understand the solution to the famous question. How many distinct sets [including the null set] can be formed from N objects. Hope this helps. --Jim Steiger = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Re: Help me an idiot
Abdul Rahman wrote: Please help me with my statistics. Question: If you order a burger from McDonald's you have a choice of the following condiments:ketchup, mustard , lettuce. pickles, and mayonnaise. A customer can ask for all thesecondiments or any subset of them when he or she orders a burger. How many different combinations of condiments can be ordered? No condiment at all conts as one combination. Your help is badly needed Just an Idiot@leftover Before you 'put yourself down' too hard, remember, ignorance can be cured, but stupid is forever. I recommend you pick the former, given a choice. So the recommended solution is a _combination_, not _permutation_. If you say that a condiment can be (a) absent or (b) present, then you have the 6*5*4*3*2*1 _permutations_ possible. For combinations, you will divide by the number of permutations for each set of a selection. For example, for the number of possible combinations of 2 items taken from the 6 possible, we would have 6C2 = 6*5*4*3*2*1/[(4*3*2*1)*(2*1)] but then you would repeat for 1, 3, 4, and 5 items selected. I don't like this - at this early hour (4:00 am local time) I sense something seriously invalid. Best go back to listing all the possibles. Keep in mind that for a permutation, the empty (absent item) slot is different, if it is empty ketchup or empty mustard. then see if the combination equation can cover them. I doubt there are 720 possible _combinations_. Jay -- Jay Warner Principal Scientist Warner Consulting, Inc. North Green Bay Road Racine, WI 53404-1216 USA Ph: (262) 634-9100 FAX: (262) 681-1133 email: [EMAIL PROTECTED] web: http://www.a2q.com The A2Q Method (tm) -- What do you want to improve today? = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Re: Help me an idiot
On 29 Apr 2001 04:09:05 GMT, [EMAIL PROTECTED] (Eric Bohlman) wrote: I wanted that with *fries* and *ketchup*! *Not* ketchup and fries! We hear you, but fries were not included in the original problem...only the 5 condiments. But you're right, fries would be good with that! Order me one. You could make this a really nice little problem by adding in fries to the no. of possible combinations. Put onion rings in the option list too. With or without a shake would be a nice touch too. But, hey, why confuse this kid any more than necessary? He's got homework to do! :-))) = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Re: Help me an idiot
I believe, the thrust of the fries reply was the overcount in the 5*4*3*. response rather than an expression of culinary preferences. - Original Message - From: J. Williams mackeral@remove~this~first~yahoo.com To: [EMAIL PROTECTED] Sent: Sunday, April 29, 2001 10:33 AM Subject: Re: Help me an idiot On 29 Apr 2001 04:09:05 GMT, [EMAIL PROTECTED] (Eric Bohlman) wrote: I wanted that with *fries* and *ketchup*! *Not* ketchup and fries! We hear you, but fries were not included in the original problem...only the 5 condiments. But you're right, fries would be good with that! Order me one. You could make this a really nice little problem by adding in fries to the no. of possible combinations. Put onion rings in the option list too. With or without a shake would be a nice touch too. But, hey, why confuse this kid any more than necessary? He's got homework to do! :-))) = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Re: Help me an idiot
On 29 Apr 2001 09:06:15 -0700, [EMAIL PROTECTED] (Zina Taran) wrote: I believe, the thrust of the fries reply was the overcount in the 5*4*3*. response rather than an expression of culinary preferences. I think he was referring to order, i.e., which item was first and then second or vice versa. Order is important in permutation problems, but is not an issue in the condiment combination problem presented in the initial post. My point was that fries were not mentioned. Had he said he wanted ketchup THEN pickles NOT pickles then ketchup all would have been right with the world. Possibly, my attempt at humor in the response eluded you. How any one likes their Big Mac is beyond the scope of this newsgroup :-)) - Original Message - From: J. Williams mackeral@remove~this~first~yahoo.com To: [EMAIL PROTECTED] Sent: Sunday, April 29, 2001 10:33 AM Subject: Re: Help me an idiot On 29 Apr 2001 04:09:05 GMT, [EMAIL PROTECTED] (Eric Bohlman) wrote: I wanted that with *fries* and *ketchup*! *Not* ketchup and fries! We hear you, but fries were not included in the original problem...only the 5 condiments. But you're right, fries would be good with that! Order me one. You could make this a really nice little problem by adding in fries to the no. of possible combinations. Put onion rings in the option list too. With or without a shake would be a nice touch too. But, hey, why confuse this kid any more than necessary? He's got homework to do! :-))) = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ = = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Help me an idiot
Please help me with my statistics. Question: If you order a burger from McDonald's you have a choice of the following condiments:ketchup, mustard , lettuce. pickles, and mayonnaise. A customer can ask for all thesecondiments or any subset of them when he or she orders a burger. How many different combinations of condiments can be ordered? No condiment at all conts as one combination. Your help is badly needed Just an Idiot@leftover = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Re: Help me an idiot
On Sat, 28 Apr 2001, Abdul Rahman wrote: Please help me with my statistics. If you order a burger from McDonald's you have a choice of the following condiments: ketchup, mustard , lettuce. pickles, and mayonnaise. A customer can ask for all these condiments or any subset of them when he or she orders a burger. How many different combinations of condiments can be ordered? No condiment at all counts as one combination. Your help is badly needed. Why? All you have to do is construct all the possibilities and count them. Shouldn't be that hard. If you want a method for dealing with more general cases, that might be another matter, of course. But even that would yield to the same procedure, if you went about it in a systematic enough fashion. So how have you approached the problem so far? (I'm a New Englander, and we tend to disapprove of laziness. If you haven't even tried to solve it yourself [and problems like this are almost certainly dealt with in your textbook!], I'm not interested in providing any help at all.) -- DFB. Donald F. Burrill [EMAIL PROTECTED] 348 Hyde Hall, Plymouth State College, [EMAIL PROTECTED] MSC #29, Plymouth, NH 03264 603-535-2597 184 Nashua Road, Bedford, NH 03110 603-472-3742 = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Re: Help me an idiot
Five different condiments, plus no condiments, means 6*5*4*3*2*1 = 720 distinct combinations. WDA end Abdul Rahman [EMAIL PROTECTED] wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... Please help me with my statistics. Question: If you order a burger from McDonald's you have a choice of the following condiments:ketchup, mustard , lettuce. pickles, and mayonnaise. A customer can ask for all thesecondiments or any subset of them when he or she orders a burger. How many different combinations of condiments can be ordered? No condiment at all conts as one combination. Your help is badly needed Just an Idiot@leftover = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Re: Help me an idiot
On Sat, 28 Apr 2001 20:35:05 GMT, W. D. Allen Sr. [EMAIL PROTECTED] wrote: Five different condiments, plus no condiments, means 6*5*4*3*2*1 = 720 distinct combinations. WDA Correct me if I am wrong, but aren't you thinking of the number of permutations, i.e., all the condiments plus the no-condiment condition included? When one takes the factorial of a number of items it is the no. of items where the ordering is important. In combinations, the order of ketchup and mustard is not an issue. He could compute the combinations of 5 items, 4 at time, then 3, etc. and add them up plus the no condiment option. To help this kid with his homework, I think a better answer might be somewhere near 32 possible combinations. McDonald's has lots of varieties or so I'm told, but not that many :-))) = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Re: Help me an idiot
W. D. Allen Sr. [EMAIL PROTECTED] wrote: Five different condiments, plus no condiments, means 6*5*4*3*2*1 = 720 distinct combinations. I wanted that with *fries* and *ketchup*! *Not* ketchup and fries! = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
