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| Combination Manhattan CAT #5 problem 31 |
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Ron Purewal
MGMAT STAFF
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for the benefit of everyone reading the forum, i've posted the solution from the practice exam website (verbatim) below.
note the way this solution proceeds: it considers all the ways in which the two daughters can sit together, and then subtracts those possibilities from the total number (because those are, after all, the possibilities that we don't want). this is why we don't explicitly have to think about one of the daughters in the front seat: if that happens, then the two daughters can't sit together. since we're figuring out the ways the daughters can set together (so that we can eventually subtract them), we don't need to enumerate those possibilities. -- original solution follows -- The easiest way to solve this question is to consider the restrictions separately. Let’s start by considering the restriction that one of the parents must drive, temporarily ignoring the restriction that the two sisters won't sit next to each other. This means that… 2 people (mother or father) could sit in the driver’s seat 4 people (remaining parent or one of the children) could sit in the front passenger seat 3 people could sit in the first back seat 2 people could sit in the second back seat 1 person could sit in the remaining back seat The total number of possible seating arrangements would be the product of these various possibilities: 2 × 4 × 3 × 2 × 1 = 48 We must subtract from these 48 possible seating arrangements the number of seating arrangements in which the daughters are sitting together. The only way for the daughters to sit next to each other is if they are both sitting in the back. This means that… 2 people (mother or father) could sit in the driver’s seat 2 people (remaining parent or son) could sit in the front passenger seat Now for the back three seats we will do something a little different. The back three seats must contain the two daughters and the remaining person (son or parent). To find out the number of arrangements in which the daughters are sitting adjacent, let’s consider the two daughters as one unit. The remaining person (son or parent) is the other unit. Now, instead of three seats to fill, we only have two "seats," or units, to fill. There are 2 × 1 = 2 ways to seat these two units. However, the daughter-daughter unit could be d1d2 or d2d1 We must consider both of these possibilities so we multiply the 2 by 2! for a total of 4 seating possibilities in the back. We could also have manually counted these possibilities: d1d2X, d2d1X, Xd1d2, Xd2d1 Now we must multiply these 4 back seat scenarios by the front seat scenarios we calculated earlier: (2 × 2) × 4 = 16 front back If we subtract these 16 "daughters-sitting-adjacent" scenarios from the total number of "parent-driving" scenarios, we get: 48 – 16 = 32 The correct answer is B. |
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| Combination Manhattan CAT #5 problem 31 |
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