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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while i understand that we can choose 4 gifts from 16 gifts in 16c4 ways, shouldnt we also include the no of ways we can choose the children, ie 4c1, then 3c1, then 2c1 and so on. Wouldnt that also be included since we are also choosing the kids to give the gifts to?
thus shouldnt it be 16c4x4c1x12c4x3c1x8c4x2c1x4c4x1?

Bunuel and KarishmaB please help

When we use 16c4, we are selecting 4 out of 16 gifts for one particular child say Child1.
Next when we use 12C4, we are selecting 4 out of 16 gifts for one particular child say Child2.
and so on...

Think about it:

Say I select A, B, C and D for child 1 using 16C4.
Then I select E, F, G, H for child 2 using 12C4
and so on.

In another instance,
I select E, F, G, H using 16C4 for child 1.
Then I select A, B, C, D for child 2 using 12C4
and so on.

These two are different instances under our 16C4*12C4*8C4*4C4 calculation.
Hence the distribution among the distinct children is already done.

Method 1 :
There are total 16 different gifts. condition is they should be distributed among 4 children equally.
Total no of ways to distribute = 16C4 x 12C4 x 8C4 x 4C4 = 16!/4!^4

Method 2 :
There are total 4 groups of gifts. It does not matter how these 4 gifts are distributed within these groups they could have any arrangement. They could be arranged in a group in 4! ways considering they are different.
now there are total 16 gifts which can be arranged in 16! ways
Total no of ways = 16!/(4! x 4! x 4! x 4!) (4! are divided 4 times because there are 4 groups and these 4 gifts can be arranged in any way within these 4 groups)