Combination 2

Question

In how many ways can 16 different gifts be divided among four children such that each child receives exactly four gifts?

16^4
(4!)^4
16!/(4!)^4
16!/4!
416

KarishmaB · Answer

Two ways to look at this:
For the first child, we choose 4 gifts in 16C4 ways.
For the second child, we choose 4 gifts in 12C4 ways.
For the third child, we choose 4 gifts in 8C4 ways.
The fourth child gets the leftover 4 gifts in 4C4 ways
Total number of ways = 16C4 * 12C4 * 8C4 *4C4
But the answers are not in this form so we need to get rid of the Cs.
16C4 * 12C4 * 8C4 * 4C4 =  \frac{16*15*14*13*12*11*10*9*8*7*6*5}{4!*4!*4!} 
Answers have 16! in them so divide and multiply the above by 4! to get:  \frac{16!}{4!*4!*4!*4!}

Another way to look at it:
Make the children stand in a row: 1 2 3 4
Arrange all the 16 gifts in front of them in 16! ways. Each child gets the 4 gifts in front of him. But these four gifts are all arranged and we don't want to arrange them. So divide 16! by 4! for each child to get  \frac{16!}{4!*4!*4!*4!}

LM · Answer

When we have N objects and N1 of the objects are alike, N2 of the rest of the objects are alike,..., and the last one Nr are alike, the total number of different permutations of these objects is:

N!/ (N1!)(N2!)....(Nr!)

Here N1 and N2 and N3 and N4 have equal number of gifts ( coincidentally) but gifts are different.

Therefore:

16!/ (4!)(4!)(4!)(4!)

Combination 2

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