If ^{{n^2} - n}{C₂}{ = ^{{n^2} - n}}{C_{10}} , then n =

English

Gujarati

Hindi

6.Permutation and Combination

easy

If $^{{n^2} - n}{C_2}{ = ^{{n^2} - n}}{C_{10}}$, then $n = $

A

$12$

B

$4$ only

C

$- 3$ only

D

$4$ or $- 3$

Solution

(d) $^{{n^2} – n}{C_2}{ = ^{{n^2} – n}}{C_{10}}{ \Rightarrow ^{{n^2} – n}}{C_{{n^2} – n – 2}}{ = ^{{n^2} – n}}{C_{10}}$

$ \Rightarrow {n^2} – n – 2 = 10$ or $n = 4,\; – 3$.

Standard 11

Mathematics

Share

Similar Questions

If for some $\mathrm{m}, \mathrm{n} ;{ }^6 \mathrm{C}_{\mathrm{m}}+2\left({ }^6 \mathrm{C}_{\mathrm{m}+1}\right)+{ }^6 \mathrm{C}_{\mathrm{m}+2}>{ }^8 \mathrm{C}_3$ and ${ }^{n-1} P_3:{ }^n P_4=1: 8$, then ${ }^n P_{m+1}+{ }^{n+1} C_m$ is equal to

hard

(JEE MAIN-2024)

In how many ways can $5$ red and $4$ white balls be drawn from a bag containing $10$ red and $8$ white balls

easy

The solution set of $^{10}{C_{x – 1}} > 2\;.{\;^{10}}{C_x}$ is

easy

All the five digits numbers in which each successive digit exceeds its predecessor are arranged in the increasing order of their magnitude. The $97^{th}$ number in the list does not contain the digit:

normal

A company has $10$ employes. The company has decided to form a team including atleast three employes and also excluding atleast three employes. Then the number of ways of forming the team is

normal