GreenlightTestPrep 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:

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

stagesStage 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 FROM MY COURSE

[you-tube]https://youtu.be/ZVkLdDVV5Bs[/you-tube]

_________________

Brent Hanneson – Creator of greenlighttestprep.com

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