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Over the course of a full-day seminar, 66 students must give short presentations. Most of the presentations are

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by BTGmoderatorDC » Sat Dec 26, 2020 11:30 pm

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Over the course of a full-day seminar, 66 students must give short presentations. Most of the presentations are independent, but Sam’s presentation refers to something in Ruth’s, so Ruth must go before Sam; similarly, Matt’s presentation refers to something in Ruth’s and something in Sam’s, so Matt’s must come after both of those. These three presentations need not be consecutive with each other: whatever the order is, Ruth’s must come before the other two but not necessarily first of the 66, and sometime later, Matt must come after the other two but not necessarily last of all. The other 63 participants can present in any order. How many orders obey these constraints?

(A) 22!

(B) 63!

(C) 63!/3!

(D) 66!/3

(E) 66!/3!



OA E

Source: Magoosh
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Source: — Problem Solving |

BTGmoderatorDC wrote:
Sat Dec 26, 2020 11:30 pm
Over the course of a full-day seminar, 66 students must give short presentations. Most of the presentations are independent, but Sam’s presentation refers to something in Ruth’s, so Ruth must go before Sam; similarly, Matt’s presentation refers to something in Ruth’s and something in Sam’s, so Matt’s must come after both of those. These three presentations need not be consecutive with each other: whatever the order is, Ruth’s must come before the other two but not necessarily first of the 66, and sometime later, Matt must come after the other two but not necessarily last of all. The other 63 participants can present in any order. How many orders obey these constraints?

(A) 22!

(B) 63!

(C) 63!/3!

(D) 66!/3

(E) 66!/3!



OA E

Source: Magoosh
First, fix those 3 students in the required order

\( \_ R \_ S \_ M \_ \)

Now, we can choose \(1\) place out of \(4\) for the next student (say A) i.e \(4C1\) ways

\( \_ A \_ R \_ S \_ M \_ \)

For \(5\)th student, we can choose from \(5\) places i.e \(5C1\) ways

Similarly, for \(66\)th student \(66C1\) ways

Therefore, total ways \(= 4C1 \cdot 5C1 \ldots 66C1 = \dfrac{66!}{3!}\)
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