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08 Apr 2015, 05:37
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| FROM Veritas Prep Blog: When Permutations & Combinations and Data Sufficiency Come Together on the GMAT! |
 While discussing Permutations and Combinations many months back, we worked through several examples of arranging people in seats. Today we bring you an interesting question based on those concepts. It brings to the fore the tricky nature of both Data Sufficiency and Combinatorics – so much so that when the two get together, it is unlimited fun! In some circumstances, we suggest you to travel the whole nine yards – i.e. solve for the answer in Data Sufficiency questions too even if you feel that sufficiency has already been established. This is especially true for quadratic equations which we assume will give us two values of x but might actually give just a single unique value (such that both roots are the same). In Combinatorics too, sometimes what may look like two distinct cases could actually give the same answer. Let’s jump on to the question. Question 1: There are x children and y chairs in a room where x and y are prime numbers. In how many ways can the x children be seated in the y chairs (assuming that each chair can seat exactly one child)? Statement 1: x + y = 12 Statement 2: There are more chairs than children. Solution: There are x children and y chairs. x and y are prime numbers. Statement 1: x + y = 12 Since x and y are prime numbers, a quick run on 2, 3, 5 shows that there are two possible cases: Case 1: x=5 and y=7 There are 5 children and 7 chairs. Case 2: x=7 and y=5 There are 7 children and 5 chairs At first glance, they might look like two different cases and you might feel that statement 1 is not sufficient alone. But note that the question doesn’t ask you for number of children or number of chairs. It asks you about the number of arrangements. Case 1: x=5 and y=7 If there are 5 children and 7 chairs, we select 5 chairs out of the 7 in 7C5 ways. We can now arrange 5 children in 5 seats in 5! ways. Total number of arrangements would be 7C5 * 5! Case 2: x = 7 and y = 5 If there are 7 children and 5 chairs, we select 5 children out of the 7 in 7C5 ways. We can now arrange 5 children in 5 seats in 5! ways. Total number of arrangements would be 7C5 * 5! Note that in both cases the number of arrangements is 7C5*5!. Combinatorics does not distinguish between people and things. 7 children on 5 seats is the same as 5 children on 7 seats because in each case you have to select 5 out of 7 (either seats or children) and then arrange 5 children in 5! ways. So actually this statement alone is sufficient! Most people would not have seen that coming! Statement 2: There are more chairs than people. We don’t know how many children or chairs there are. This statement alone is not sufficient. Answer: A We were tempted to answer the question as (C) but it was way too easy. Statement 1 gave 2 cases and statement 2 narrowed it down to 1. Be aware that if it looks too easy, you are probably missing something! Now, what if we alter the question slightly and make it: Question 2: There are x children and y chairs arranged in a circle in a room where x and y are prime numbers. In how many ways can the x children be seated in the y chairs (assuming that each chair can seat exactly one child)? Statement 1: x + y = 12 Statement 2: There are more chairs than children. Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog! |
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08 Apr 2015, 06:39
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Question 2: There are x children and y chairs arranged in a circle in a room where x and y are prime numbers. In how many ways can the x children be seated in the y chairs (assuming that each chair can seat exactly one child)?
Statement 1: x + y = 12
case 1 : x=7, y=5 ----> \(^7C_5 * (5-1)!\)
case 2: x=5, y=7 ----> as there are more chairs than there are children we cannot use circular permutation . there will be two empty chairs.
Choose 2 chairs which are empty out of 7 , #ways \(^7C_2\) and now the 5 children can be arranged in 5! ways .
note \(^7C_2\) = \(^7C_5\)
so total number of ways are \(^7C_5\) * \(5!\)
Not Sufficient.
Statement 2: There are more chairs than children
Insufficient.
together A and B.
we know this is case:2 of statement A .should be sufficient.
answer C
VeritasPrepKarishma please confirm whether this is correct .
Statement 1: x + y = 12
case 1 : x=7, y=5 ----> \(^7C_5 * (5-1)!\)
case 2: x=5, y=7 ----> as there are more chairs than there are children we cannot use circular permutation . there will be two empty chairs.
Choose 2 chairs which are empty out of 7 , #ways \(^7C_2\) and now the 5 children can be arranged in 5! ways .
note \(^7C_2\) = \(^7C_5\)
so total number of ways are \(^7C_5\) * \(5!\)
Not Sufficient.
Statement 2: There are more chairs than children
Insufficient.
together A and B.
we know this is case:2 of statement A .should be sufficient.
answer C
VeritasPrepKarishma please confirm whether this is correct .
29 Apr 2015, 21:28
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Lucky2783
Question 2: There are x children and y chairs arranged in a circle in a room where x and y are prime numbers. In how many ways can the x children be seated in the y chairs (assuming that each chair can seat exactly one child)?
Statement 1: x + y = 12
case 1 : x=7, y=5 ----> \(^7C_5 * (5-1)!\)
case 2: x=5, y=7 ----> as there are more chairs than there are children we cannot use circular permutation . there will be two empty chairs.
Choose 2 chairs which are empty out of 7 , #ways \(^7C_2\) and now the 5 children can be arranged in 5! ways .
note \(^7C_2\) = \(^7C_5\)
so total number of ways are \(^7C_5\) * \(5!\)
Not Sufficient.
Statement 2: There are more chairs than children
Insufficient.
together A and B.
we know this is case:2 of statement A .should be sufficient.
answer C
VeritasPrepKarishma please confirm whether this is correct .
Statement 1: x + y = 12
case 1 : x=7, y=5 ----> \(^7C_5 * (5-1)!\)
case 2: x=5, y=7 ----> as there are more chairs than there are children we cannot use circular permutation . there will be two empty chairs.
Choose 2 chairs which are empty out of 7 , #ways \(^7C_2\) and now the 5 children can be arranged in 5! ways .
note \(^7C_2\) = \(^7C_5\)
so total number of ways are \(^7C_5\) * \(5!\)
Not Sufficient.
Statement 2: There are more chairs than children
Insufficient.
together A and B.
we know this is case:2 of statement A .should be sufficient.
answer C
VeritasPrepKarishma please confirm whether this is correct .
Your answer is correct but your case 2 has a problem. If you have 7 distinct objects, then you can choose 5 in 7C5 ways. But here, the chairs arranged in a circle are identical initially. Say you choose two chairs right next to each other, how are they any different from any other two chairs right next to each other?
When you have 5 children and 7 chairs in a circle, you place the first child on any chair in 1 way.
Then you have 4 children and 6 distinct chairs. You can place the 4 children in 6*5*4*3 ways.
Total number of ways of placing 5 children in 7 chairs in a circle = 6*5*4*3
This is not the same as case 1 (7C5*4!).
So you need to know whether the number of chairs is more or the number of children is more. Hence you need both statement to answer the question.
Answer (C)
