Permutations & Combinations
Selecting numbers in AP
Grade 11
Question:
<p>Number of ways in which three numbers can be chosen from \(\{1, 2, 3, \ldots, n\}\) such that they are in AP and the number of integers between the first and the second chosen is equal to the number of integers between the second and the third is</p>
<p>(a) \(\binom{n-2}{3}\)</p>
<p>(b) \(\binom{n-3}{3} + \binom{n-3}{2}\)</p>
<p>(c) \(\frac{(n-2)(n-3)(n-4)}{6}\)</p>
<p>(d) \(\binom{n}{2}\)</p>
Step-by-Step Solution
Key Concept: Transform the problem into counting positive integral solutions by introducing auxiliary variables for gaps between chosen numbers.
<p>Let $a_0$ be the number of objects to the left of the first object chosen, $a_1$ be the number of objects between the first and the second, $a_2$ be the number of objects between the second and the third and $a_3$ be the number of objects to the right of the third object. Then:</p><p>$a_0, a_3 \geq 0$ and $a_1, a_2 \geq 1$</p><p>Also $a_0 + a_1 + a_2 + a_3 = n - 3$</p><p>Let $a = a_0 + 1, b = a_3 + 1$, then $a \geq 1, b \geq 1$ such that $a + a_1 + a_2 + b = n - 1$</p><p>The total number of positive integral solutions of this equation is $\binom{n-1-1}{4-1} = \binom{n-2}{3} = \binom{n-3}{3} + \binom{n-3}{2} = \frac{(n-2)(n-3)(n-4)}{6}$</p>
Correct Answer: a, b, c