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In how many different ways can 6 identical belts and 5 identical hats
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07 Oct 2019, 08:08
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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, 09:34
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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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Updated on: 06 Dec 2020, 06:45
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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: 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)
Re: In how many different ways can 6 identical belts and 5 identical hats
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04 Nov 2019, 20:44
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Expert Reply
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
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05 Dec 2020, 23:09
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