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In how many ways can 6 identical coins be distributed among Alex, Bea,

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In how many ways can 6 identical coins be distributed among Alex, Bea,  [#permalink]

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New post 02 Jul 2017, 22:13
1
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A
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Question Stats:

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In how many ways can 6 identical coins be distributed among Alex, Bea, and Chad? Note: Some people may receive zero coins.

(A) 15
(B) 21
(C) 28
(D) 30
(E) 36

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In how many ways can 6 identical coins be distributed among Alex, Bea,  [#permalink]

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New post 02 Jul 2017, 23:09
Since some people can receive zero coins,
there are 3 ways of distributing the coins when 2 of them receive zero coins
A,B,C
6,0,0
0,6,0
0,0,6

Total : 3 ways

When one of them receives zero coin, the other two can receive {(1,5),(2.4)} coins
there are 6 ways of distributing each of these coins among A,B and C
A,B,C
1,5,0
5,1,0
1,0,5
5,0,1
0,1,5
0,5,1
This same pattern re-occurs for number (2,4)

When one of them receives zero coin, the other two can receive (3,3) coins
but there are 3 ways of distributing each of these coins among A,B and C
A,B,C
3,3,0
0,3,3
3,0,3

Total : 12 + 3 = 15 ways

When neither of them receives zero coins, they can be distributed in 2 ways.
(1,1,4) in 3 ways and (1,2,3) in 3!(6) ways

Total : 9 ways

There is one way of arranging the coins such that each of them has 2 coins.
(2,2,2)

Total : 1 way

Hence we have a total of 3+15+9+1 = 28 ways(Option C)
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In how many ways can 6 identical coins be distributed among Alex, Bea,  [#permalink]

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New post 03 Jul 2017, 00:59
saswata4s wrote:
In how many ways can 6 identical coins be distributed among Alex, Bea, and Chad? Note: Some people may receive zero coins.

(A) 15
(B) 21
(C) 28
(D) 30
(E) 36


Is there a shorter and faster way to solve this, rather than counting each scenario?


Thanks,
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Re: In how many ways can 6 identical coins be distributed among Alex, Bea,  [#permalink]

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New post 27 Sep 2017, 01:10
Novice90 wrote:
saswata4s wrote:
In how many ways can 6 identical coins be distributed among Alex, Bea, and Chad? Note: Some people may receive zero coins.

(A) 15
(B) 21
(C) 28
(D) 30
(E) 36


Is there a shorter and faster way to solve this, rather than counting each scenario?


Thanks,
Novice


I'm not sure if there is a faster solution, but this is what I did and it is similar to the solution provided by pushpitkc

Some people can receive zero coins, we have 2 cases:
1. 2 guys receive zero coins, then the distribution is to be calculated for (6,0,0) = \(\frac{3!}{2!}\)= 3
2. 1 guy recieves zero coins, then the distribution is calculated as follows
(1,5,0) = \(3!\)= 6
(2,4,0) = \(3!\)= 6
(3,3,0) = \(\frac{3!}{2!}\)= 3

All receive atleast one coin
(1,1,4) = \(\frac{3!}{2!}\)= 3
(1,2,3) = \(3!\)= 6
(2,2,2) = \(\frac{3!}{3!}\)= 1

So, the total is 3 + 6 + 6 + 3 + 3 + 6 + 1 = 28
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Re: In how many ways can 6 identical coins be distributed among Alex, Bea,  [#permalink]

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New post 27 Sep 2017, 03:47
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1
saswata4s wrote:
In how many ways can 6 identical coins be distributed among Alex, Bea, and Chad? Note: Some people may receive zero coins.

(A) 15
(B) 21
(C) 28
(D) 30
(E) 36


You will need to enumerate here though you can start with an obvious starting point and proceed logically: Start with the max number of coins you can give first and then the next highest number and then the last. The second number should never be more than the first and the third number should never be more than the second.

(6, 0, 0) - The 6 coins can be given to the 3 people in 3 ways

(5, 1, 0) - The 3 different number of coins can be distributed among 3 people in 3! = 6 ways.

(4, 2, 0) - The 3 different number of coins can be distributed among 3 people in 3! = 6 ways.

(4, 1, 1) - The 4 coins can be given to the 3 people in 3 ways

(3, 3, 0) - The 0 coins can be given to the 3 people in 3 ways

(3, 2, 1) - The 3 different number of coins can be distributed among 3 people in 3! = 6 ways.

(2, 2, 2) - There is 1 way of distributing 2 coins each

Total = 3 + 6 + 6 + 3 + 3 + 6 + 1 = 28

Answer (C)
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Re: In how many ways can 6 identical coins be distributed among Alex, Bea,  [#permalink]

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New post 28 Sep 2017, 03:49
VeritasPrepKarishma wrote:
saswata4s wrote:
In how many ways can 6 identical coins be distributed among Alex, Bea, and Chad? Note: Some people may receive zero coins.

(A) 15
(B) 21
(C) 28
(D) 30
(E) 36


You will need to enumerate here though you can start with an obvious starting point and proceed logically: Start with the max number of coins you can give first and then the next highest number and then the last. The second number should never be more than the first and the third number should never be more than the second.

(6, 0, 0) - The 6 coins can be given to the 3 people in 3 ways

(5, 1, 0) - The 3 different number of coins can be distributed among 3 people in 3! = 6 ways.

(4, 2, 0) - The 3 different number of coins can be distributed among 3 people in 3! = 6 ways.

(4, 1, 1) - The 4 coins can be given to the 3 people in 3 ways

(3, 3, 0) - The 0 coins can be given to the 3 people in 3 ways

(3, 2, 1) - The 3 different number of coins can be distributed among 3 people in 3! = 6 ways.

(2, 2, 2) - There is 1 way of distributing 2 coins each

Total = 3 + 6 + 6 + 3 + 3 + 6 + 1 = 28

Answer (C)


Responding to a pm:

Quote:
This question does not seem different from the one posted in "https://gmatclub.com/forum/in-how-many-ways-5-different-chocolates-be-distributed-to-4-children-231187.html#p1781064" .

Can you please enlighten on why the same approach is not applicable in current problem.
########################

in how many ways 5 different chocolates be distributed to 4 children such that any child can get any number of chocolates?

1) 20
2) 24
3) 120
4) 625
5) 1024

I saw this question on a youtube video and the solution showed 4*4*4*4*4 = 1024, but i was thinking it would be 5*5*5*5 and this answer isn't even in the options. Please help me understand this.


Let the 5 cholcolates be : C1C1 , C2C2 , C3C3 , C4C4 & C5C5
And there be 4 Students : S1S1 , S2S2, S3S3 & S4S4

Now, The first Chocolate can be given in 4 ways ( Either to S1S1 or S2S2 or S3S3 or S4S4 )
The Second Chocolate can be given in 4 ways ( Either to S1S1 or S2S2 or S3S3 or S4S4 )
The Third Chocolate can be given in 4 ways ( Either to S1S1 or S2S2 or S3S3 or S4S4 )
The Fourth Chocolate can be given in 4 ways ( Either to S1S1 or S2S2 or S3S3 or S4S4 )
The fifth Chocolate can be given in 4 ways ( Either to S1S1 or S2S2 or S3S3 or S4S4 )

Hence, the total No of ways possible is 4545 = 10241024

Hence, answer will be (E) 1024...


The two questions are not same.
In the original question, the coins are identical. In the other question, the chocolates are all different.

These posts discuss such variations:
https://www.veritasprep.com/blog/2011/1 ... inatorics-–-part-1/
https://www.veritasprep.com/blog/2011/1 ... s-part-ii/
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Re: In how many ways can 6 identical coins be distributed among Alex, Bea,  [#permalink]

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New post 14 Dec 2017, 00:32
the secret formula for this math problem is (total number of groups)! / (the number of same groups)!
this is an example. AAABBCC. There are 7! / (3!2!2!)
For the problem, (6, 0, 0) will have 3 groups, 2 groups are same.
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Re: In how many ways can 6 identical coins be distributed among Alex, Bea, &nbs [#permalink] 14 Dec 2017, 00:32
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In how many ways can 6 identical coins be distributed among Alex, Bea,

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