Problem Solving

A deviant from the problem https://www.beatthegmat.com/combo-t73840.html
an excellent question from BTG practice questions.

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

A) 16^4
B) (4!)^4
C) 16!/[(4!)^4]
D) 16!/4!
E) 4^16


OA is C

Anurag,

I did this method. Is there another approach to this? I tried applying the "stick method" used by another Gurome expert. I dont think it works well.


[quote="Anurag@Gurome"][quote="arora007"]In how many ways can 16 different gifts be divided among four children such that each child receives exactly four gifts?

A) 16^4
B) (4!)^4
C) 16!/[(4!)^4]
D) 16!/4!
E) 4^16[/quote]

The 1st children choose 4 gifts out of 16 in 16[b]C[/b]4 ways.
The 2nd children choose 4 gifts out of remaining 12 in 12[b]C[/b]4 ways.
The 3rd children choose 4 gifts out of remaining 8 in 8[b]C[/b]4 ways.
The 4th children choose 4 gifts out of remaining 4 in 4[b]C[/b]4 ways.

Hence, total number of ways = (16[b]C[/b]4)*(12[b]C[/b]4)*(8[b]C[/b]4)*(4[b]C[/b]4)

Let's rearrange this to get answer in the given form.

(16[b]C[/b]4)*(12[b]C[/b]4)*(8[b]C[/b]4)*(4[b]C[/b]4)
= [16!/(12!*4!)]*[12!/(8!*4!)]*[8!/(4!*4!)]*1
= 16!/(4!*4!*4!*4!)
= 16!/[(4!)^4]

[spoiler]The correct answer is C.[/spoiler][/quote]

Hi Anurag thanks for your response!

I guess I have one very basic and fundamental question.

Well... I too on the test had tried in a similar manner, except that I took a + instead of a * inbtween the terms...

(16C4) + (12C4) + (8C4)+ (4C4)

I seem to get confused when we have which sign.. is there some mechanism with which I can be sure of ?

Permutations/Combinations & at times Probability is something which I cannot at times get hold of...

Also Any special tips how I can make my 49 a 51 on G-Day?

Took a slightly different approach with anagramming:

16 gifts, each child (A, B, C and D) receiving 4 of them:

so anagram AAAABBBBCCCCDDDD.

Therefore combination is 16!/(4!4!4!4!), or 16!/[(4!)^4], so answer C

Anurag@Gurome wrote:

arora007 wrote:In how many ways can 16 different gifts be divided among four children such that each child receives exactly four gifts?

A) 16^4
B) (4!)^4
C) 16!/[(4!)^4]
D) 16!/4!
E) 4^16

The 1st children choose 4 gifts out of 16 in 16C4 ways.
The 2nd children choose 4 gifts out of remaining 12 in 12C4 ways.
The 3rd children choose 4 gifts out of remaining 8 in 8C4 ways.
The 4th children choose 4 gifts out of remaining 4 in 4C4 ways.

Hence, total number of ways = (16C4)*(12C4)*(8C4)*(4C4)

Let's rearrange this to get answer in the given form.

(16C4)*(12C4)*(8C4)*(4C4)
= [16!/(12!*4!)]*[12!/(8!*4!)]*[8!/(4!*4!)]*1
= 16!/(4!*4!*4!*4!)
= 16!/[(4!)^4]

The correct answer is C.

Anurag,
Would you mind explaining a bit more how you get from
= [16!/(12!*4!)]*[12!/(8!*4!)]*[8!/(4!*4!)]*1
= 16!/(4!*4!*4!*4!)
??

I'm a bit of a rookie at this combinatorics questions, so would really appreciate a more detailed explanation.

Thanks

tgou008 wrote:
Anurag,
Would you mind explaining a bit more how you get from
= [16!/(12!*4!)]*[12!/(8!*4!)]*[8!/(4!*4!)]*1
= 16!/(4!*4!*4!*4!)
??

I'm a bit of a rookie at this combinatorics questions, so would really appreciate a more detailed explanation.

Thanks

Notice that we have 12! and 8! in the numerator and denominator when we multiply all the terms. The 12! is in the denominator in the first term, but in the numerator in the second. The 8! is in the denominator in the second term and numerator in the 3rd term

Therefore, those amounts will cancel out, leaving us with the 16! and four 4!s

arora007 wrote: A) 16^4
B) (4!)^4
C) 16!/[(4!)^4]
D) 16!/4!
E) 4^16

I might find it helpful if I work in reverse on some of these more difficult combination problems. So I'm going to try and figure out the description for each answer and see if something matches up to the question at hand.

A) 16^4 would apply if we had 16 gifts and 4 kids choosing 1 gift (with replacement) and order does matter.
B) (4!)^4 would apply if we had only four different gifts (but 4 sets of the same 4 gifts) being distribute to 4 kids, with order of the gifts being important within each set.
C) I probably would pass on trying to explain this one, so we'll leave it as a possibility
D) 16!/4! is the same as 16P12 - so if we had to arrange 12 gifts from 16 where order does count. Doesn't really match with our problem
E) If four kids got to pick 4 times from 4 gifts (with replacement), and order does matter.

I think my elimination, we're left with C. Did I mess up any explanations?

Think of the 16 gifts arranged in a row, such that the first child gets the 4 gifts from 1 to 4, the second child gets the 4 gifts from 5 to 8, the third child gets the 4 gifts from 9 to 12 and the fourth child gets the 4 gifts from 13 to 16.

There are 16! possibilities to arrange all the gifts. Since the order in which each child receives his/her 4 gifts does not matter, we should divide 16! by 4!x4!x4!x4! For example, a child getting the gifts A,B,C,D was counted 4! times and not just once, because in 16! we took into account all the possibilities to arrange those 4 gifts in a row. And this we have done for each child, therefore 4! appears for times in the denominator, once for each child.

So, the correct answer is C

Nice question. This is IIT-JEE level question!!!

i just saw the video from an email i got from beatthegmat.

it went something like 16C4, you get the first 4 products of 16! and divide it by 4!

so 16c4 * 12c4 * 8c4 * 4c4 = 16*15*14*13/4! * 12*11*10*9/4! * 8*7*6*5/4! * 4*3*2*1/4! = 16!/4!^4

sorry if my post is bad my first time trying these problems on the forum

Got stuck at 16C4 and did not go on to 12C4*8C4 etc

Easy looking but really testing one. Good one.

(C) 16!/[(4!)^4]

arora007 wrote:Hi Anurag thanks for your response!

I guess I have one very basic and fundamental question.

Well... I too on the test had tried in a similar manner, except that I took a + instead of a * inbtween the terms...

(16C4) + (12C4) + (8C4)+ (4C4)

I seem to get confused when we have which sign.. is there some mechanism with which I can be sure of ?

Permutations/Combinations & at times Probability is something which I cannot at times get hold of...

Also Any special tips how I can make my 49 a 51 on G-Day?

Keep it simple
OR means +
AND means *

Here it is AND so *