nCr+1nCr=21(given)⇒(r+1)!(n−r−1)!n!r!(n−r)!n!=21⇒r!(n−r)!(r+1)!(n−r−1)!=21⇒r!(n−r)(n−r−1)!(r+1)⋅r!(n−r−1)!=21⇒n−r=2(r+1)⇒n−r=2r+2⇒n=3r+2..(i) Again, nCr+2nCr+1=32⇒(r+2)!(n−r−2)!n!(r+1)!(n−r−1)!n!=32⇒(r+1)!(n−r−1)!(r+2)!(n−r−2)!=32⇒(r+1)!(n−r−1)(n−r−2)!(r+2)(r+1)!(n−r−2)!=322(n−r−1)=3(r+2)⇒2n−2r−2=3r+6⇒2n=5r+8...(ii)Multiplyequation(i)by22 n=6 r+4
...(iii)Fromequations(ii)and(iii)⇒6r+4r=5r+8=4Puttingthevalueofr
inequation(ii)2n2n2nn=5(4)+8=20+8=28=14.$\begin{aligned} & { }^n \mathrm{P}_r:{ }^n \mathrm{C}_r=\frac{{ }^n \mathrm{P}_r}{{ }^n \mathrm{C}_r} \\ & \begin{aligned} \frac{\frac{n!}{(n-r)!}}{\frac{n!}{r!(n-r)!}} & =\frac{n!}{(n-r)!} \times \frac{r!(n-r)!}{n!} \\ & =r!=4! \\ & =4 \times 3 \times 2 \times 1=24 .\end{aligned} \quad[\because r=4]\end{aligned}$
