Question

42 nCr-1 = 10, nCr= 3 nCr+1 Find n and r​

Answer 1

Step-by-step explanation:

Given, \begin{gathered}^nC_{r-1}=36\\\\$^nC_r=84\\\\$^nC_{r+1}=126\end{gathered}

∴ \frac{^nC_r}{^nC_{r-1}}=\frac{84}{36}nCr−1nCr=3684

⇒{n!/r!(n - r)!}/{n!/(r-1)!(n-r-1)!} = 7/3

⇒(n - r + 1)/r = 7/3

⇒3n - 3r + 3 = 7r

⇒3n + 3 = 10r ------(1)

Again, \frac{^nC_{r+1}}{^nC_r}=\frac{126}{84}nCrnCr+1=84126

⇒{n!/(r+1)!(n-r-1)!}/{n!/r!(n-r)!} = 3/2

⇒(n - r)/(r + 1) = 3/2

⇒2n - 2r = 3r + 3

⇒2n - 3 = 5r ------(2)

From equations (1) and (2),

3n + 3 = 4n - 6

⇒9 = n and r = 3

Now, rC₂ = ³C₂ = 3!/2! = 3

Hence, answer is 3