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

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In how many different ways can 6 identical belts and 5 identical hats  [#permalink]

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New post 07 Oct 2019, 09:08
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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

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In how many different ways can 6 identical belts and 5 identical hats  [#permalink]

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New post 07 Oct 2019, 10:34
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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




GMATPrepNow wrote:
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
GMAT Club Legend
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Joined: 12 Sep 2015
Posts: 4076
Location: Canada
In how many different ways can 6 identical belts and 5 identical hats  [#permalink]

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New post 07 Oct 2019, 17:34
Top Contributor
GMATPrepNow wrote:
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


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:
Image
ASIDE: When it comes to populating the matrix, the key piece of information is that the question tells us that each child receives at least 1 item, which means there are ZERO children in the bottom right box (indicating those children who received neither a hat nor a belt)

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

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Re: In how many different ways can 6 identical belts and 5 identical hats  [#permalink]

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New post 04 Nov 2019, 21:44
1
GMATPrepNow wrote:
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


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)
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Re: In how many different ways can 6 identical belts and 5 identical hats   [#permalink] 04 Nov 2019, 21:44
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