Manjareev wrote:Find the total number of ways in which 35 identical marbles can be distributed among 5 boys such that each boy gets odd number of marbles?
1) 39C4
2) 19C4
3) 15C4
4) 20C4
5) 25C4
Pls explain the answer.
An odd value can be represented as 2x + 1.
Since each of the 5 boys receives an ODD number of marbles, and the total number of marbles is 35, we can represent the sum as follows:
(2a + 1) + (2b + 1) + (2c + 1) + (2d + 1) + (2e + 1) = 35.
2(a+b+c+d+e) + 5 = 35
a+b+c+d+e = 15.
Now the question becomes:
How many ways can 15 identical marbles be distributed among 5 children a, b, c, d and e?
Use the SEPARATOR method.
15 identical marbles are to be separated into -- at most -- 5 groupings.
Thus, we need 15 marbles and 4 separators:
OOO|OO|OOOOO|OO|OOO
Each arrangement of the elements above represents one way to distribute the 15 marbles among 5 children a, b, c, d and e:
OOOO|OOO|OOO|OO|OO = a gets 4 marbles, b gets 3 marbles, c gets 3 marbles, d gets 2 marbles, e gets 2 marbles.
OOOOOOOO||||OOOOOOO = a gets 8 marbles, e gets 7 marbles.
OOOOOOOOOOOOOOO|||| = a gets all 15 marbles.
And so on.
To count all of the possible distributions, we simply need to count the number of ways to arrange the 19 elements above (the 15 identical marbles and the 4 identical separators).
The number of ways to arrange 19 elements = 19!.
But when an arrangement includes identical elements, we must divide by the number of ways to arrange the identical elements.
The reason:
When the identical elements swap positions, the arrangement doesn't change, reducing the total number of unique arrangements.
Thus, we must divide by 15! (the number of ways to arrange the 15 identical marbles) and 4! (the number of ways to arrange the 4 identical separators):
19!/(15! 4!) = = (19*18*17*16)/4! = 19C4.
The correct answer is
B.
Last edited by GMATGuruNY on Sun Sep 23, 2012 2:21 pm, edited 1 time in total.
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