In how many different ways can 6 identical belts and 5 identical hats

Question

In how many different ways can 6 identical belts and 5 identical hats be distributed among 8 different children, 
so that each child receives at least 1 item, no child receives 2 or more belts, and no child receives 2 or more hats?

A) 240
B) 256
C) 420
D) 480
E) 560

nick1816 · Answer

Tricky one!!

3 children will get 1 hat and 1 belt.
3 children will get a belt.
2 children will get a hat.

Total number of ways= 8C3*5C3*2C2= 56*10*1=560

BrentGMATPrepNow · Answer

The first thing we need to do is determine how many children fall into each category (e.g., receive a hat but no belt, receive a belt but no hat, or receive both a hat and a belt)

To do so, we can use the   Double Matrix Method  . This technique can be used for most questions featuring a population in which each member has two characteristics associated with it (aka overlapping sets questions). 
Here, we have a population of children, and the two characteristics are: 
- receives a hat or doesn't receive a hat
- receives a belt or doesn't receive a belt

When we apply the Double Matrix Method, the distribution of the 8 children looks like this:
 https://i.imgur.com/c3OqSGn.png

ASIDE: Let me explain how I populated the matrix. 
The 8 represents the 8 children. 
Since the top two boxes represent children who received a belt, we know that the SUM of those two boxes must be 6 (since there are 6 belts to distribute)
If there are 8 children and 6 receive a belt, then 2 children do not receive belts. 
Since the bottom two boxes represent children who didn't receive a belt, we know that the SUM of those two boxes must be 2

The same applies to the columns. 
The two boxes on the left side represent children who received a hat. Since there are 5 hats to be distributed, the sum of those two boxes must be 5. 
If there are 8 children and 5 receive a hat, then 3 children do not receive hats. 
Since the two right-most boxes represent children who didn't receive a hat, we know that the SUM of those two boxes must be 3

Finally, since we're told that   each child receives at least 1 item  , this means there are ZERO children in the bottom right box (indicating those children who received neither a hat nor a belt)
At this point we can complete the rest of the matrix

If you want to learn more about the Double Matrix Method, watch the video below. 
 
Okay, once we've determined the number of children who fall into each category, it's simply a matter of choosing the children for each category. 
We'll do so in  stages

Stage 1 : Select 3 children to receive both a hat and a belt 
Since the order in which we select the children does not matter, we can use combinations. 
We can select 3 children from 8 children in 8C3 ways (56 ways)
So, we can complete stage 1 in  56  ways

Stage 2 : Select 2 children to receive a hat but no belt 
There are now 5 children remaining.
Once again, we'll use combinations (since the order in which we select the children does not matter) 
We can select 2 children from the remaining 5 children in 5C2 ways (10 ways)
So, we can complete stage 2 in  10  ways.

Stage 3 : Select 3 children to receive a belt but no hat
There are now 3 children remaining.
We can select 3 children from the remaining 3 children in 3C3 ways (1 way)
So, we can complete stage 3 in  1  way.

By the Fundamental Counting Principle (FCP), we can complete all 3 stages (and thus distribute all of the hats and belts) in  (56)(10)(1)  ways (= 560 ways)

Answer: E

Cheers, 
Brent

RELATED VIDEO 
 https://www.youtube.com/watch?v=jK-tiBrrf04

KarishmaB · Answer

Responding to a pm:
Each child must get one item and no child should receive the same item twice. Note that we have 11 items and 8 children. This means 3 children will get 2 items each - one belt and one hat. 
Of the 8, let's select the 3 kids who will get both in 8C3 ways.
Now 5 kids are left. Of these 5, select 2 to give them the leftover 2 hats in 5C2 ways.
The leftover 3 kids will just get the 3 leftover belts in 1 way.

Total ways = 8C3 * 5C2 * 1 = 560

Answer (E)

hadimadi · Answer

Hi,

here is how I have done it (no guarantee):

First, we check how many ways we can distribute 6B and 5H so that every of the 8 children has exactly one item. This means that the possible combinations of H and B are, with LH, LB denoting "Left Hats, Left Belts" after we distribute for exactly 8 kids:

B H LB LH

Case 1 6 2 0 3

Case 2 5 3 1 2

Case 3 4 2 2 1

Case 4 3 5 3 0

Let's go case by case and check first how many combinations we have for exactly 1 item per child, and how the left hats and left belts can then be distributed:

C1: A total of 8!/ =28. We have 3LB, but only the 6 children with a Belt can get them. We can therefore pick 6!/ =20 different children. So after distributing the leftovers, we have 28*20=560 possibilities.

C2: Repeating the process as in C1, but now have 3*(560) possibilities.

C3: Repeating as in C1, but now have 3*560 possibilities.

C4: Repeating as in C1, and have 560 possibilities.

Counting all of them together, we have 8*560 total possibilities. But we have to remove similar combinations.
We realize that every single of these cases has 3 kids with a hat and a belt, 3 with exactly 1 belt, and 2 with exactly 1 hat, meaning that every single one of the cases also distributes all of our 6B and 5H among the same 11 children. This means that the amount of total different possibilities has to be min(C1,C2,C3,C4)=560.

Fdambro294 · Answer

Got a little lucky with this one (if the answer choices were different….who knows).

Listed out the 8 children and started putting pen to paper:

C1 - C2 - C3 - C4 - C5 - C6 - C7 - C8

Then, just LIST OUT one of the possible scenarios. Since we have 6 belts and 5 hats, we have 11 items.

This means that 3 children must, at the very least, get more than one item. They can not get 2 hats or 2 belts. It must be belt and hat.

C1, C2, C3: Belt - Hat

C4, C5: Hat

C6, C7, C8: Belt

At this point, in which we arrange the items like this:

1st) choose which 3 of the 8 children will get the (belt - hat) combo

(8 c 3) = 56

2nd) arrange the 5 remaining items: 2 identical hats AND 3 identical belts

5! / (3! 2!) = 10 ways

(56) (10) = 560

No need to check whether any other ways can work. This is the highest possible answer.

*E*
560

Note: since the belts are identical and the hats are identical, it does NOT matter which belt or which hat a child receives.

All the matters is whether he or she has a belt, a hat, or both a belt and a hat.

EDIT:
After seeing the solution, the “path to the answer” involves seeing that, under the conditions, the above is the ONLY way to distribute the items.

Posted from my mobile device

szcz · Answer

This question does not mention that all the hats and belts must be used. We can distribute the hats and belts in such a way that only 8 hats + belts will be used and 3 hats+belts will remain surplus. Or we may use 9 and so on. This increases the number of cases exponentially. I think the question needs to mention that all the hats and belts will be used. Experts kindly check.  Bunuel   KarishmaB   yashikaaggarwal   bb   IanStewart

KarishmaB · Answer

"In how many different ways can 6 identical belts and 5 identical hats be distributed among 8 different children..."

We have to distribute 6 belts and 5 hats. This means we cannot distribute only say 8 items or 9 items etc. We must distribute all 11.

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In how many different ways can 6 identical belts and 5 identical hats

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