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23 Mar 2024, 01:54
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unicornilove
Hi Bunuel, what am I missing here? My answer is 6C2. Choose 2 slots out of 6
D= donut ; Slot = S
S1 D S2 D S3 D S4 D S5 D S6
We'll get combinations like: \(dd|d|dd\) 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 \({n+r-1}_C_{r-1}\).
In our case we'll get: \({n+r-1}_C_{r-1}={5+3-1}_C_{3-1}={7}C2=\frac{7!}{5!2!}=21\).
Similar question: https://gmatclub.com/forum/integers-les ... ml#p710836
D= donut ; Slot = S
S1 D S2 D S3 D S4 D S5 D S6
Bunuel
rxs0005
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 \(\frac{7!}{5!2!}=21\).(A) 21
(B) 42
(C) 120
(D) 504
(E) 5040
We'll get combinations like: \(dd|d|dd\) 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 \({n+r-1}_C_{r-1}\).
In our case we'll get: \({n+r-1}_C_{r-1}={5+3-1}_C_{3-1}={7}C2=\frac{7!}{5!2!}=21\).
Similar question: https://gmatclub.com/forum/integers-les ... ml#p710836
Not following your method. The correct one is provided in my post. Here are more questions to learn the stars and bars method from:
https://gmatclub.com/forum/12-marbles-a ... l#p3353907
https://gmatclub.com/forum/in-how-many- ... l#p3352713
https://gmatclub.com/forum/in-how-many- ... l#p3335084
18 Apr 2024, 23:13
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Bunuel
rxs0005
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 \(\frac{7!}{5!2!}=21\).(A) 21
(B) 42
(C) 120
(D) 504
(E) 5040
We'll get combinations like: \(dd|d|dd\) 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 \({n+r-1}_C_{r-1}\).
In our case we'll get: \({n+r-1}_C_{r-1}={5+3-1}_C_{3-1}={7}C2=\frac{7!}{5!2!}=21\).
Similar question: https://gmatclub.com/forum/integers-les ... ml#p710836
We're sharing 5 donuts among 3 individuals here right, and we don't have any constraint as to who will get any number of donuts.
So, each donut will have 3 options to go to (3 individuals). So, shouldn't the answer be 3x3x3x3x3 = 3^5.
The method you've stated is very valid, but my method is from questions like "3 letters need to be posted into 4 letter boxes". Where each letter has 4 options to go into, so 4^3. I don't seem to find any difference between these questions.
If you think the method I stated doesn't belong here pls clarify it. GMATNinja KarishmaB. Anyone pls clear my doubt.
18 Apr 2024, 23:40
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Sahith_Manikanta
Bunuel
rxs0005
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 \(\frac{7!}{5!2!}=21\).(A) 21
(B) 42
(C) 120
(D) 504
(E) 5040
We'll get combinations like: \(dd|d|dd\) 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 \({n+r-1}_C_{r-1}\).
In our case we'll get: \({n+r-1}_C_{r-1}={5+3-1}_C_{3-1}={7}C2=\frac{7!}{5!2!}=21\).
Similar question: https://gmatclub.com/forum/integers-les ... ml#p710836
We're sharing 5 donuts among 3 individuals here right, and we don't have any constraint as to who will get any number of donuts.
So, each donut will have 3 options to go to (3 individuals). So, shouldn't the answer be 3x3x3x3x3 = 3^5.
The method you've stated is very valid, but my method is from questions like "3 letters need to be posted into 4 letter boxes". Where each letter has 4 options to go into, so 4^3. I don't seem to find any difference between these questions.
If you think the method I stated doesn't belong here pls clarify it. GMATNinja KarishmaB. Anyone pls clear my doubt.
The two situations are different. You are talking about distributing distinct objects - 3 different letters. Here this question involves distributing 5 identical objects (items of a given food are considered identical unless stated otherwise. So 5 apples means identical apples).
The handling of distinct objects vs identical objects is very very different. This question will be solves using separators as Bunuel mentioned above.
I have discussed various such cases in my section on Distributions in Combinations study module. You can check it out on Sunday through Super Sundays program. Details here:
https://youtu.be/gN_vlDpUflo
