Consider a plygon of sides `'n'` which satisflies the equation `3. .^(n)p_(4) = .^(n-1)p_(5)`
Number of quadrilaterals that can be made using the vertices of the polygon of sides `'n'` if exactly two adjacent side of the quadrilateral are common to the sides of polygon is

Options

A. `50`
B. `60`
C. `70`
D. `.^(10)C_(3)-70`

Answer

A, D

Explanation

`3.(n!)/((n - 4)!) = ((n - 1)!)/((n - 6)!), 3n = (n - 4)(n - 5)`
`rArr 3n = n^(2) - 9n + 20 rArr n^(2) - 12n + 20 = 0`
`n = 10, n = 2` , not possible
so `n = 10`

`n = 10`
Consecutive two sides can be formed by `(1, 2, 3)` point `4, 10` can't be selected but one point out of ramain `5` point can be.
selected in `.^(5)C_(1)` ways
So total quadritaterals `= 10x^(5)C_(1) = 50`