The three integers \(n_1\), \(n_2\) and \(n_3\) satisfy \(0 < n_1 < n_2 < n_3\) and \(n_1 + n_2 > n_3\). Find the number of ways of choosing the pair of numbers \(n_1\) and \(n_2\) in the cases \(n_3 = 9\) and \(n_3 = 10\).
Given that \(n_3 = 2n + 1\), where \(n\) is a positive integer, write down an expression (which you need not prove is correct) for the number of ways of choosing the pair of numbers \(n_1\) and \(n_2\). Simplify your expression.
Write down and simplify the corresponding expression when \(n_3 = 2n\), where \(n\) is a positive integer.
You have \(N\) rods, of lengths \(1, 2, 3, \ldots, N\) (one rod of each length). You take the rod of length \(N\), and choose two more rods at random from the remainder, each choice of two being equally likely. Show that, in the case \(N = 2n + 1\) where \(n\) is a positive integer, the probability that these three rods can form a triangle (of non-zero area) is
$$\frac{n - 1}{2n - 1}.$$
Find the corresponding probability in the case \(N = 2n\), where \(n\) is a positive integer.
You have \(2M + 1\) rods, of lengths \(1, 2, 3, \ldots, 2M + 1\) (one rod of each length), where \(M\) is a positive integer. You choose three at random, each choice of three being equally likely. Show that the probability that the rods can form a triangle (of non-zero area) is
$$\frac{(4M + 1)(M - 1)}{2(2M + 1)(2M - 1)}.$$
Note: \(\sum_{k=1}^{K} k^2 = \frac{1}{6}K(K + 1)(2K + 1)\).
Solution:
If \(n_3 = 9\) and we are looking for \(0 < n_1 < n_2 < n_3\) we can consider values for each \(n_2\).
\begin{array}{clc|c}
n_2 & \text{range} & \text{count} \\ \hline
6 & 4-5 & 2 \\
7 & 3-6 & 4 \\
8 & 2-7 & 6 \\ \hline
& & 12
\end{array}
When \(n_3 = 10\)
\begin{array}{clc|c}
n_2 & \text{range} & \text{count} \\ \hline
6 & 5 & 1 \\
7 & 4-6 & 3 \\
8 & 3-7 & 5 \\
9 & 2-8 & 7 \\ \hline
& & 16
\end{array}
When \(n_3 = 2n+1\) we can have \(2 + 4 + \cdots + 2n-2 = n(n-1)\)
When \(n_3 = 2n\) we can have \(1 + 3 + \cdots + 2n-3 = (n-1)^2\)
For the 3 rods to form a triangle, it suffices for the sum of the lengths of the shorter rods to be larger than \(N\). When \(N = 2n+1\) there are \(n(n-1)\) ways this can happen, out of \(\binom{2n}{2}\) ways to choos the numbers, ie
\begin{align*}
&& P &= \frac{n(n-1)}{\frac{2n(2n-1)}{2}} \\
&&&= \frac{n-1}{2n-1}
\end{align*}
When \(N = 2n\) there are \((n-1)^2\) ways this can happen, out of \(\binom{2n-1}{2}\) ways, ie
\begin{align*}
&& P &= \frac{(n-1)^2}{\frac{(2n-1)(2n-2)}{2}} \\
&&&= \frac{n-1}{2n-1}
\end{align*}
The number of ways this can happen is:
\begin{align*}
C &= \sum_{k=3}^{2M+1} \# \{ \text{triangles where }k\text{ is largest} \} \\
&= \sum_{k=1}^{M} \# \{ \text{triangles where }2k+1\text{ is largest} \} +\sum_{k=1}^{M} \# \{ \text{triangles where }2k\text{ is largest} \}\\
&= \sum_{k=1}^{M} n(n-1)+\sum_{k=1}^{M} (n-1)^2\\
&= \sum_{k=1}^{M} (2n^2-3n+1)\\
&= \frac26M(M+1)(2M+1) - \frac32M(M+1) + M \\
&= \frac16 M(4M+1)(M-1)
\end{align*}
Therefore the probability is
\begin{align*}
&& P &= \frac{M(4M+1)(M-1)}{6 \binom{2M+1}{3}} \\
&&&= \frac{M(4M+1)(M-1)}{(2M+1)2M(2M-1)} \\
&&&= \frac{(4M+1)(M-1)}{2(2M+1)(2M-1)}
\end{align*}
The random variable \(X\) has the probability density function on the interval \([0, 1]\):
$$f(x) = \begin{cases}
nx^{n-1} & 0 \leq x \leq 1, \\
0 & \text{elsewhere},
\end{cases}$$
where \(n\) is an integer greater than 1.
Let \(\mu = E(X)\). Find an expression for \(\mu\) in terms of \(n\), and show that the variance, \(\sigma^2\), of \(X\) is given by
$$\sigma^2 = \frac{n}{(n + 1)^2(n + 2)}.$$
In the case \(n = 2\), show without using decimal approximations that the interquartile range is less than \(2\sigma\).
Write down the first three terms and the \((k + 1)\)th term (where \(0 \leq k \leq n\)) of the binomial expansion of \((1 + x)^n\) in ascending powers of \(x\).
By setting \(x = \frac{1}{n}\), show that \(\mu\) is less than the median and greater than the lower quartile.
Note: You may assume that
$$1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \cdots < 4.$$