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Chapter 1: Problem 2

In how many ways can 15 (identical) candy bars be distributed among five children so that the youngest gets only one or two of them?

Short Answer

Expert verified
The total number of ways 15 identical candy bars can be distributed among five children such that the youngest gets only one or two of them is the sum of \(\binom{18}{4}\) and \(\binom{17}{4}\)

Step by step solution

01

Distributing Candy Bars - Scenario 1

In the first scenario, the youngest child gets one candy bar. This leaves us with 14 candy bars to distribute among five children. According to the stars and bars method, the number of ways to distribute \(n\) identical items among \(r\) recipients is given by \(\binom{n+r-1}{r-1}\). In this case, \(n = 14\) and \(r = 5\), which gives us \(\binom{14+5-1}{5-1} = \binom{18}{4}\).
02

Distributing Candy Bars - Scenario 2

In the second scenario, the youngest child gets two candy bars. This leaves us with 13 candy bars to distribute among five children. Likewise, using the stars and bars method, the number of ways is \(\binom{13+5-1}{5-1} = \binom{17}{4}\).
03

Sum both scenarios

To find the total number of ways the candy bars can be distributed, we need to add the number of ways from both scenarios because they are mutually exclusive (the youngest child can't get both one and two candy bars at the same time). So we sum \(\binom{18}{4}\) and \(\binom{17}{4}\)

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Most popular questions from this chapter

a) In how many ways can 17 be written as a sum of 2 's and 3 's if the order of the summands is (i) not relevant? (ii) relevant? b) Answer part (a) for 18 in place of 17 .

Show that for all positive integers \(m\) and \(n\), $$ n\left(\begin{array}{c} m+n \\ m \end{array}\right)=(m+1)\left(\begin{array}{l} m+n \\ m+1 \end{array}\right) $$.

There are \(b_{4}(=14)\) ways to arrange \(1,2,3, \ldots, 8\) in two rows of four so that (1) the integers increase in value as each row is read, from left to right, and (2) in any column the smaller integer is on top. Find, as in part (d) of Example \(1.43\), a) the arrangements that correspond to each of the following. i) 10110010 ii) 11001010 iii) 11101000 b) the lists of four 1 's and four 0 's that correspond to each of these arrangements of \(1,2,3, \ldots, 8\). i) 1345 2678 ii) 1237 4568 iii) 1245

a) If \(n\) and \(r\) are positive integers with \(n \geq r\), how many solutions are there to $$ x_{1}+x_{2}+\cdots+x_{r}=n $$ where each \(x_{i}\) is a positive integer, for \(1 \leq i \leq r ?\) b) In how many ways can a positive integer \(n\) be written as a sum of \(r\) positive integer summands ( \(1 \leq r \leqq n)\) if the order of the summands is relevant?

a) For \(n \geq 4\), consider the strings made up of \(n\) bits - that is, a total of \(n 0\) 's and 1's. In particular, consider those strings where there are (exactly) two occurrences of 01 . For example, if \(n=6\) we want to include strings such as 010010 and 100101, but not 101111 or 010101 . How many such strings are there? b) For \(n \geq 6\), how many strings of \(n 0\) 's and 1 's contain (exactly) three occurrences of 01 ? c) Provide a combinatorial proof for the following: For \(n \geq 1\), $$ 2^{n}=\left(\begin{array}{c} n+1 \\ 1 \end{array}\right)+\left(\begin{array}{c} n+1 \\ 3 \end{array}\right)+\cdots+ \begin{cases}\left(\begin{array}{c} n+1 \\ n \end{array}\right), & n \text { odd } \\ \left(\begin{array}{c} n+1 \\ n+1 \end{array}\right), & n \text { even. }\end{cases} $$
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