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Chapter 8: Problem 5

In how many ways can one distribute ten distinct prizes among four students with exactly two students getting nothing? How many ways have at least two students getting nothing?

Short Answer

Expert verified
There are 6144 ways to distribute ten distinct prizes among four students with exactly two students getting nothing, and 6149 ways when at least two students get nothing.

Step by step solution

01

Choose the students who will receive prizes

Since two students are not receiving any prizes, we must select two students out of four who will. Let's use combinations to find the number of ways. Using the choose function, this is \( C(4,2) = 6 \) ways.
02

Distribute prizes among chosen students

Each prize is distinct, so we can distribute them among two students in a number of ways. Given there are 10 prizes, each prize can either go to Student 1 or Student 2. This gives us \( 2^{10} \) ways.
03

Multiply to find total

Since these are independent events, we multiply the results of the two steps. Therefore, there are \( 6*2^{10} = 6144 \) ways to distribute the prizes with exactly two students getting nothing.
04

The case where at least two students receive nothing

We need to calculate separately for when two, three or all four students receive no prizes. We have already calculated for when two students receive nothing which is 6144 ways. If three students receive nothing, there would be \( C(4,1) = 4 \) ways to choose a student and \( 1^{10} \) ways to distribute prizes, hence \( 4*1^{10} = 4 \) ways. If all four students receive nothing, there would be 1 way which is no distribution. Therefore, summing for all cases, there are \( 6144+4+1 = 6149 \) ways where at least two students receiving nothing.

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Most popular questions from this chapter

Ms. Pezzulo teaches geometry and then biology to a class of 12 advanced students in a classroom that has only 12 desks. In how many ways can she assign the students to these desks so that (a) no student is seated at the same desk for both classes? (b) there are exactly six students each of whom occupies the same desk for both classes?

Let \(n \in \mathbf{Z}^{+}\), (a) Determine \(\phi\left(2^{n}\right)\). (b) Determine \(\phi\left(2^{n} p\right)\), where \(p\) is an odd prime.

If 13 cards are dealt from a standard deck of 52, what is the probability that these 13 cards include (a) at least one card from each suit? (b) exactly one void (for example, no clubs)? (c) exactly two voids?

. For which \(n \in \mathbf{Z}^{+}\)is \(\phi(n)\) odd?

Give a combinatorial argument to verify that for all \(n \in \mathbf{Z}^{+}\), $$ n !=\left(\begin{array}{l} n \\ 0 \end{array}\right) d_{0}+\left(\begin{array}{l} n \\ 1 \end{array}\right) d_{1}+\left(\begin{array}{l} n \\ 2 \end{array}\right) d_{2}+\cdots+\left(\begin{array}{l} n \\ n \end{array}\right) d_{n}=\sum_{k=0}^{n}\left(\begin{array}{l} n \\ k \end{array}\right) d_{k} $$ (For each \(1 \leq k \leq n, d_{k}=\) the number of derangements of 1 \(2,3, \ldots, k ; d_{0}=1 .\) )
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