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Larry, Michael, and Doug have five donuts to share. If any [#permalink]
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04 Feb 2011, 16:05
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Larry, Michael, and Doug have five donuts to share. If any one of the men can be given any whole number of donuts from 0 to 5, in how many different ways can the donuts be distributed? (A) 21 (B) 42 (C) 120 (D) 504 (E) 5040
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Re: Combinations tough [#permalink]
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rxs0005 wrote: Larry, Michael, and Doug have five donuts to share. If any one of the men can be given any whole number of donuts from 0 to 5, in how many different ways can the donuts be distributed?
(A) 21 (B) 42 (C) 120 (D) 504 (E) 5040 Consider this: we have 5 donuts \(d\) and 2 separators \(\), like: \(ddddd\). How many permutations (arrangements) of these symbols are possible? Total of 7 symbols (5+2=7), out of which 5 \(d\)'s and 2 \(\)'s are identical, so \(\frac{7!}{5!2!}=21\). We'll get combinations like: \(ddddd\) this would mean that Larry got 2 donuts, Michael got 1 donut and Doug got 2 donuts, so to the left of the first separator are Larry's donuts, between the separators are Michael's donuts and to the right of the second separator are Doug's donuts Answer: A. This can be done with direct formula as well: The total number of ways of dividing n identical items (5 donuts in our case) among r persons or objects (3 persons in our case), each one of whom, can receive 0, 1, 2 or more items (from zero to 5 in our case) is \({n+r1}_C_{r1}\). In our case we'll get: \({n+r1}_C_{r1}={5+31}_C_{31}={7}C2=\frac{7!}{5!2!}=21\). Similar question: integerslessthan85291.html#p710836
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Re: Combinations tough [#permalink]
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04 Feb 2011, 18:46



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Larry, Michael, and Doug have five donuts to share. If any one o [#permalink]
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17 Jan 2013, 09:17
Another awesome variation to this question (the one mentioned above) would be:
How many integer solutions (x,y,z) are there to the equation: x+y+z = 20, where x is at least equal to 3, y is at least equal to 4, and z is at least equal to 5



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Re: Larry, Michael, and Doug have five donuts to share. If any [#permalink]
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17 Jan 2013, 11:56
should it not state identical donuts to share? relatively straightforward question but got confused as to whether we needed to use combinations formula and then arrange between the 3 people



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Re: Combinations tough [#permalink]
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03 Sep 2013, 08:05
Bunuel wrote: rxs0005 wrote: Larry, Michael, and Doug have five donuts to share. If any one of the men can be given any whole number of donuts from 0 to 5, in how many different ways can the donuts be distributed?
(A) 21 (B) 42 (C) 120 (D) 504 (E) 5040 Consider this: we have 5 donuts \(d\) and 2 separators \(\), like: \(ddddd\). How many permutations (arrangements) of these symbols are possible? Total of 7 symbols (5+2=7), out of which 5 \(d\)'s and 2 \(\)'s are identical, so \(\frac{7!}{5!2!}=21\). We'll get combinations like: \(ddddd\) this would mean that Larry got 2 donuts, Michael got 1 donut and Doug got 2 donuts, so to the left of the first separator are Larry's donuts, between the separators are Michael's donuts and to the right of the second separator are Doug's donuts Answer: A. This can be done with direct formula as well: The total number of ways of dividing n identical items (5 donuts in our case) among r persons or objects (3 persons in our case), each one of whom, can receive 0, 1, 2 or more items (from zero to 5 in our case) is \({n+r1}_C_{r1}\). In our case we'll get: \({n+r1}_C_{r1}={5+31}_C_{31}={7}C2=\frac{7!}{5!2!}=21\). Similar question: integerslessthan85291.html#p710836Bunuel  could you explain how this problem would sound if I used a simple counting principle like Larry can get 5 doughnuts Michael 4...all the way to have 5! and therefore I would get 5!= 120, which would obviously be too easy but I am a little confused as to the difference in wording. Solution and those dividers make good sense, so thanks for that.
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Re: Larry, Michael, and Doug have five donuts to share. If any [#permalink]
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05 Mar 2014, 11:51
If you get confused about combinations, there's a simple way to count these combinations as well, by counting the number of ways 5 can be summed with 3 numbers.
{5,0,0} = 3 possibilities. {4,1,0} = 6 possibilities. {3,2,0} = 6 possibilities. {3,1,1} = 3 possibilities. {2,2,1} = 3 possibilities. Total = 21 possibilities.
Tip: For each set, we only have to consider numbers less than the first; for instance, we wouldn't consider {2,3,0} because that's already accounted for in a permutation of {3,2,0}



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Re: Larry, Michael, and Doug have five donuts to share. If any [#permalink]
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03 Jul 2014, 09:07
Dear Bunuel, Thanks for the great explanation. But there is a catch that is not clear to me. I read all other similar types of questions and also went through the explanations given by you. There you have used 3 separators for all the cases but here you have used 2. Could you tell me how would I know that how many separators should I use. Thanks in advance.



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Re: Larry, Michael, and Doug have five donuts to share. If any [#permalink]
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05 Jul 2014, 07:09
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Re: Larry, Michael, and Doug have five donuts to share. If any [#permalink]
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06 Jul 2014, 05:50
Why 3^5 is not a correct answer? Considering each donut has 3 possibilities(L,M &D)... Kindly regards



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Re: Larry, Michael, and Doug have five donuts to share. If any [#permalink]
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06 Jul 2014, 10:29



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Re: Larry, Michael, and Doug have five donuts to share. If any [#permalink]
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19 Nov 2014, 09:59
Keep in mind that 2 people might not even get any donuts at all.. so total 7! and 5 donuts are identical and 2 not given are identical .. so 7!/ 5! 2! = 21



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Re: Larry, Michael, and Doug have five donuts to share. If any [#permalink]
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20 Dec 2014, 10:53
L + M + D = 5 As L,M,D >= 0 We need to distribute 5 donots and 2 empty vessels. So 7!/(5!*2!) or 7C2 = 21  A)
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Re: Larry, Michael, and Doug have five donuts to share. If any [#permalink]
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Re: Larry, Michael, and Doug have five donuts to share. If any [#permalink]
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13 Jan 2016, 02:33
This problem similar to:
In how many ways can 3 distinct rings be worn on 4 fingers? (each finger can contain 03 rings)
OR
In how many ways can 3 distinct letters be distributed among 4 letterboxes? (each letterbox can contain 03 letters)
The solution to this problem is given as 4^3, as you have 3 slots (each slot representing a ring or a letter) and each slot can have 4 possibilities
4 4 4 = 4^3
Yet, the 3^5 does not work in the case of the donut. Please explain.



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