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In how many ways can 6 identical coins be distributed among Alex, Bea, [#permalink]
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]
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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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.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/1 ... inatorics-–-part-1/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/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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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, [#permalink]
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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


Number of ways of dividing 'n' identical objects into 'r' groups such that each group can contain any number of objects is given by

\(n+r-1_C_{r-1}\)

So,
The number of ways of dividing 6 identical coins among the 3 is

\(6+3-1_C_{3-1}\)

= \(8_{C_2}\)

= \(\frac{8!}{6!*2!}\)

= \(28\)

Hence C is the correct answer.
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Re: In how many ways can 6 identical coins be distributed among Alex, Bea, [#permalink]
I like to think of these questions graphically

Let say the 3 people are A, B and C

A: X number of coins
B: Y number of coins
C: 6 - X - Y number of coins

The only constraint is that X + Y <= 6

If you plot that inequality on the graph and see the total number of points in the permissible region, you will get 28 as your answer
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Re: In how many ways can 6 identical coins be distributed among Alex, Bea, [#permalink]
Asked: In how many ways can 6 identical coins be distributed among Alex, Bea, and Chad? Note: Some people may receive zero coins.

CCCCCC||

There are 6 identical coins are 2 partitions
Number of ways = 8!/6!2! = 28

IMO C
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Re: In how many ways can 6 identical coins be distributed among Alex, Bea, [#permalink]
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There are 2 important equations for these types of arrangements


If the number of non negative integer solutions for the equation \(x_1 + x_2 \space + \space ..+ \space x_n \space = \space n\), then the number of ways the distribution can be done is:


(i) \(^{n+r−1}C_{r−1}\). In this case, value of any variable can be zero.


(ii) \(^{n−1}C_{r−1}\). In this case, minimum value for any variable is 1.




Given that: n = 6 coins, r = 3 people and and some people may get 0 coins (as there is no statement saying each should get at least 1 coin.)

Therefore the total number of ways = \(^{n + r - 1}C_{r-1} = \space ^{6 + 3 - 1}C_{3 - 1} = \space ^8C_2 = \frac{8 \space * \space 7 }{2 \space * \space 1} = 28\)


Option C

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