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In how many different ways can 6 identical belts and 5 identical hats
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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?
In how many different ways can 6 identical belts and 5 identical hats
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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?
In how many different ways can 6 identical belts and 5 identical hats
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07 Oct 2019, 17:34
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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:
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)
Re: In how many different ways can 6 identical belts and 5 identical hats
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04 Nov 2019, 21:44
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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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