r wanted to distribute 5 candies to each of her 10 friends.
How many candies did she totally distribute?
Answers
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GMAT Club Forum Index Problem Solving (PS)
In how many ways can 5 different candiesbe distributed among : Problem Solving (PS)
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voodoochild
Oct 21, 2012
00:01ABCDE
DIFFICULTY: 45% (medium) QUESTION STATS: based on 229 sessions
57% (01:06) correct
43% (01:36) wrong
In how many ways can 5 different candies be distributed among four children? (Children could get 0 candies or more than one candy)
(A) 4^5
(B) 5^4
(C) 5!
(D) 4!
(E) 4!*5!
Method1 (Long method):
N= the child doesn't get anything
5-N-N-N : 4!/3! * 5C5 = 4
4-1-N-N : 4!/2! * 5C4 *1C1 = 5*12 = 60
3-2-N-N : 4!/2! * 5C3*2C2 = 12*10 =120
3-1-1-N : 4!/2! * 5C3*2C2 = 12*10 = 100
2-2-1-N : 4!/2! * 5C2*3C2*1c1 *(1/2) (1/2 because we are double counting 2-2) = 12 * 15 = 180
2-1-1-1 : 4!/3! * 5C2*3C3 = 4*10=40
If I add these numbers, it doesn't equal to 1024. What's my mistake?
Thanks
Spoiler: OA
Kudos
1 kudos, 16 bookmarks
Most Helpful Community Reply
shrikar23
Sep 2, 2013
there are total 5 candies. Lets say A,B,C,D and E.
Candy A can be distributed among any 4 children. So there are 4 ways of distributing candy A.
similarly candy B,C,D and E can be distributed in 4 ways.
So total ways of distribution is
4*4*4*4*4 = 4^5.
A
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General Discussion
EvaJager
Oct 22, 2012
voodoochild wrote:
In how many ways can 5 different candies be distributed among four children? (Children could get 0 candies or more than one candy)
(A) 4^5
(B) 5^4
(C) 5!
(D) 4!
(E) 4!*5!
Method1 (Long method):
N= the child doesn't get anything
5-N-N-N : 4!/3! * 5C5 = 4
4-1-N-N : 4!/2! * 5C4 *1C1 = 5*12 = 60
3-2-N-N : 4!/2! * 5C3*2C2 = 12*10 =120
3-1-1-N : 4!/2! * 5C3*2C2 = 12*10 = 100
2-2-1-N : 4!/2! * 5C2*3C2*1c1 *(1/2) (1/2 because we are double counting 2-2) = 12 * 15 = 180
2-1-1-1 : 4!/3! * 5C2*3C3 = 4*10=40
If I add these numbers, it doesn't equal to 1024. What's my mistake?
Thanks
5-N-N-N : 4!/3! * 5C5 = 4 - OK
4-1-N-N : 4!/2! * 5C4 *1C1 = 5*12 = 60 - OK, divide by 2! because the two zero's (N) are indistinguishable
3-2-N-N : 4!/2! * 5C3*2C2 = 12*10 =120 - OK
3-1-1-N : 4!/2! * 5C3*2C2 = 12*10 = 100 - it should be 4! * 5C3 = 24*10 = 240, you should not divide by 2!, each ball is distinct
2-2-1-N : 4!/2! * 5C2*3C2*1C1 *(1/2) (1/2 because we are double counting 2-2) = 12 * 15 = 180 - NO - it should be 4! * 5C2 * 3C2 * 1C1/2! = 360, doesn't matter which group of 2 you have chosen first, but later, when permuting them, it counts
2-1-1-1 : 4!/3! * 5C2*3C3 = 4*10=40 - NO - it should be 4! * 5C2 = 240, all the balls are distinct, so you should not divide by 3!.
The total is 4 + 60 + 120 + 240 + 360 + 240 = 1024.
But why sweat all the way along????
Kudos
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voodoochild
Oct 22, 2012
EvaJager wrote:
But why sweat all the way along????
4 word answer : "No pain, no gain" --- something that my physical trainer taught me a few years ago. :)
Kudos
BN1989
Oct 31, 2012
EvaJager wrote:
voodoochild wrote:
In how many ways can 5 different candies be distributed among four children? (Children could get 0 candies or more than one candy)
(A) 4^5
(B) 5^4
(C) 5!
(D) 4!
(E) 4!*5!
Method1 (Long method):
N= the child doesn't get anything
5-N-N-N : 4!/3! * 5C5 = 4
4-1-N-N : 4!/2! * 5C4 *1C1 = 5*12 = 60
3-2-N-N : 4!/2! * 5C3*2C2 = 12*10 =120
3-1-1-N : 4!/2! * 5C3*2C2 = 12*10 = 100
2-2-1-N : 4!/2! * 5C2*3C2*1c1 *(1/2) (1/2 because we are double counting 2-2) = 12 * 15 = 180
2-1-1-1 : 4!/3! * 5C2*3C3 = 4*10=40
Answer:
50 candies
Step-by-step explanation:
She has to give 5 candies to each friend.
There are 10 friends.
So, 10 *5 = 50
So the total number of candies she distributed are 50 candies