In how many ways can 16 different gifts be divided among four children

Total 16 different Gifts, and 4 children.
Thus any one child gets 16C4 gifts,
then the other child gets 12C4 gifts(16 total - 4 already given),
then the third one gets 8C4 gifts,
and the last child gets 4C4 gifts.
Since order in which each child gets the gift is not imp, thus, ans :
16C4 * 12C4 * 8C4 * 4C4 = 16! / (4!)^4
Ans : C.

16 gifts can be distributed to 4 children in 16C4 ways.
Remaining 12 gifts can be distributed to 4 children in 12C4 ways.
Remaining 8 gifts can be distributed to 4 children in 8C4 ways.
Lastly, remaining 4 gifts can be distributed to 4 children in 4C4 ways.

Total ways = 16C4 * 12C4 * 8C4 * 4C4
= 16!/(4!)^4

Hence option (C).
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help, VeritasPrepKarishma
why the formula (ax)! / a! (x!)^a does not work?

Note that you are distributing 16 gifts among 4 children. The children are distinct. If you were distributing the gifts among 4 identical baskets such that each basket has exactly 4 gifts, then you would need to divide by 4! too.

Note where this formula comes from:

Put all 16 gifts in a row in 16! ways.


G1, G2, G3, G4, ... , G16

Now split them into 4 groups

G1, G2, G3, G4 || G5, G6, G7, G8 || ... || G13, G14, G15, G16

If the 4 groups are identical (like identical baskets), divide by 4!

16!/4!

Now since we don't want to arrange the gifts within the group, divide by 4! four times

16!/4!*(4!)^4

But since we have 4 distinct children here, we will not divide by the first 4!

The first child can choose any 4 gifts from the 16 gifts; thus (s)he has 16C4 ways to choose them. Once (s)he has chosen his or her 4 gifts, the second child can choose any 4 gifts from the remaining 12 gifts; thus (s)he has 12C4 ways to choose them. Likewise, the third child has 8C4 ways to choose his or her 4 gifts and the last child has 4C4 ways to choose his or her 4 gifts.
Thus the total number of ways the 16 gifts can be divided among the four children such that each child will receive 4 gifts is:

16C4 x 12C4 x 8C4 x 4C4

(16 x 15 x 14 x 13)/4! x (12 x 11 x 10 x 9)/4! x (8 x 7 x 6 x 5)/4! x (4 x 3 x 2 x 1)/4!

(16 x 15 x 14 x 13 x … x 4 x 3 x 2 x 1)/(4! x 4! x 4! x 4!)

16!/(4!)^4

Answer: C

METHOD-1

Gift for first child can be selected in 16C4 ways = 1820
Gift for first child can be selected in 12C4 ways = 495
Gift for first child can be selected in 8C4 ways = 70
Gift for first child can be selected in 4C4 ways = 1

Total Ways to distribute gifts = (16C4*12C4*8C4*4C4) = 16!/(4!)^4

METHOD-2

We can arrange the 16 gifts in 16! ways considering that first 4 gifts are for 1st child, next 4 gifts are for 2nd child and so on...

But since the arrangement of 4 gifts received by child is irrelevant included in 16! so we need to eliminate the effect by dividing the result by 4! four times as there are four groups of 4gifts each group

hence answer = 16!/(4!^4)


Answer: option C


In my opinion, the numbers given here are too big for GMAT to consider. This questions would have been apt if there were 4 children with 8 gifts where each child was to receive 2 gifts.

Let's say the children are named A, B, C, and D

Stage 1: Select 4 gifts to give to child A
Since the order in which we select the 4 gifts does not matter, we can use combinations.
We can select 4 gifts from 16 gifts in 16C4 ways (= 16!/(4!)(12!))
So, we can complete stage 1 in 16!/(4!)(12!) ways

Stage 2: select 4 gifts to give to child B
There are now 12 gifts remaining
Since the order in which we select the 4 gifts does not matter, we can use combinations.
We can select 4 gifts from 12 gifts in 12C4 ways (= 12!/(4!)(8!))
So, we can complete stage 2 in 12!/(4!)(8!) ways


Stage 3: select 4 gifts to give to child C
There are now 8 gifts remaining
We can select 4 gifts from 8 gifts in 8C4 ways (= 8!/(4!)(4!))
So, we can complete stage 3 in 8!/(4!)(4!) ways

Stage 4: select 4 gifts to give to child D
There are now 4 gifts remaining
NOTE: There's only 1 way to select 4 gifts from 4 gifts, but if we want the answer to look like the official answer, let's do the following:
We can select 4 gifts from 4 gifts in 4C4 ways (= 4!/4!)
So, we can complete stage 4 in 4!/4! ways

By the Fundamental Counting Principle (FCP), we can complete all 4 stages (and thus distribute all 16 gifts) in [16!/(4!)(12!)][12!/(4!)(8!)][8!/(4!)(4!)][4!/4!] ways

A BUNCH of terms cancel out to give us = 16!/(4!)⁴

Answer: C

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

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

Number of ways to distribute gifts = 16C4 × 12X4 × 8C4 × 4C4 = 16!/4!12! × 12!/4!8! × 8!/4!4! × 4!/4!0! = 16!/4!4!4!4! = 16!/(4!)^4

IMO C

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