First slide

Combinations

Question

Let $Q_{n}$ be the number of possible quadrilaterals formed by joining vertices of an n sided regular polygon. If $Q_{n + 1} - Q_{n} = 20$ then value of $n$ is:

Moderate

A

8
B

7
C

6
D

5

Solution

$Q_{n} =$ Number of ways of choosing four vertices out of $n$
$\begin{array}{l} =^{n} C_{4} \\ ∴ 20 = Q_{n + 1} - Q_{n} =^{n + 1} C_{4} -^{n} C_{4} \\ =^{n} C_{4} +^{n} C_{3} -^{n} C_{4} =^{n} C_{3} \\ [∵^{n} C_{r - 1} +^{n} C_{r} =^{n + 1} C_{r}] \\ \frac{1}{6} n (n - 1) (n - 2) = 20 \end{array}$
$\begin{array}{ll} \Rightarrow n^{3} - 3 n^{2} + 2 n - 120 = 0 \\ \Rightarrow n^{3} - 6 n^{2} + 3 n^{2} - 18 n + 20 n - 120 = 0 \\ \Rightarrow (n - 6) (n^{2} + 3 n + 20) = 0 \\ As n \in N, n = 6 \end{array}$

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