The number of ways of choosing triplet `(x , y ,z)` such that `zgeqmax{x, y}a n dx ,y ,z in {1,2, n ,n+1}` is a. `^n+1C_3+^(n+2)C_3` b. `n(n+1)(2n+1)//6` c. `1^2+2^2++n^2` d. `2((^(n+2)C_3))_(-^(n+2))C_2`

Options

A. `.^(n+1)C_(2)+.^(n+2)C_(2)`
B. `1/6n(n+1)(2n+1)`
C. `1^(2)+2^(2)+…+n^(2)`
D. `2^(n+2)C_(3)- .^(n+1)C_(2)`

Answer

A, B, C, D

Explanation

If `z=1`
so `x,y=1`
`xyrarr{1,2}`
number of ways `=2^(2)`
`z=3`
`x,yepsilon{1,2,3}`
number of ways `=3^(2)`
Similarly `n^(2)`
`1^(2)+2^(2)……….n^(2)`