Home
Class 12
MATHS
The number of ways of choosing triplet (...

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`

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)`

Text Solution

Verified by Experts

The correct Answer is:
A, B, C, D
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)`
Promotional Banner

Topper's Solved these Questions

  • PERMUTATIONS AND COMBINATIONS

    ARIHANT MATHS|Exercise Exercise (Passage Based Questions)|15 Videos

  • PERMUTATIONS AND COMBINATIONS

    ARIHANT MATHS|Exercise Exercise (Single Integer Answer Type Questions)|10 Videos

  • PERMUTATIONS AND COMBINATIONS

    ARIHANT MATHS|Exercise Exercise (Single Option Correct Type Questions)|30 Videos

  • PARABOLA

    ARIHANT MATHS|Exercise Exercise (Questions Asked In Previous 13 Years Exam)|35 Videos

  • PROBABILITY

    ARIHANT MATHS|Exercise Exercise (Questions Asked In Previous 13 Years Exam)|55 Videos

Similar Questions

Explore conceptually related problems

Determine n if "^(2n)C_3 : "^nC_3 = 11 : 1

If "^(2n)C_3 : ^nC_2= 12: 1 , find n .

Determine n if "^(2n)C_3 : "^nC_3 =12:1

The number of real negative terms in the binomial expansion of (1+i x)^(4n-2),n in N ,x >0 is n b. n+1 c. n-1 d. 2n

The total number of ways in which 2n persons can be divided into n couples is a. (2n !)/(n ! n !) b. (2n !)/((2!)^3) c. (2n !)/(n !(2!)^n) d. none of these

The number of positive integers satisfying the inequality C(n+1,n-2) - C(n+1,n-1)<=100 is

The number of functions f: {1,2, 3,... n}-> {2016, 2017} , where ne N, which satisfy thecondition f1)+f(2)+ ...+ f(n) is an odd number are a. 2^n b. n*2^(n-1) c. 2^(n-1) d. n!

Number of ways in which three numbers in A.P. can be selected from 1,2,3,..., n is a. ((n-1)/2)^2 if n is even b. ((n-2)/4)^ if n is even c. ((n-1)/4)^2 if n is odd d. none of these

Prove that ""^(n)C_(3)+""^(n)C_(7) + ""^(n)C_(11) + ...= 1/2{2^(n-1) - 2^(n//2 )sin"" (npi)/(4)}