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

Answer

A, B, C, D

Explanation

Triplets with
(i) x=y`ltz`
(ii) `x lt y lt z`
(iii) `y lt x lt z`
can be chosen in `.^(n+1)C_(2),.^(n+1)C_(3),.^(n+1)C_(3)` ways.
`therefore.^(n+1)C_(2)+.^(n+1)C_(3)+.^(n+1)C_(3)=.^(n+2)C_(3)+.^(n+1)C_(3)`
`=2(.^(n+2)C_(3))-.^(n+1)C_(2)`
`=(n(n-1)(2n+1))/(6)`