Binomial Theorem
Combinatorics + Permutation/Combination
nta_pyq_2024_jan
Grade 11
Question:
If for some $m,n$; ${}^6C_m + 2({}^6C_{m+1}) + {}^6C_{m+2} > {}^8C_3$ and ${}^{n-1}P_3 : {}^nP_4 = 1:8$, then ${}^nP_{m+1} + {}^{n+1}C_m$ is equal to
Step-by-Step Solution
Key Concept: Use Pascal's identity twice: ${}^6C_m+2{}^6C_{m+1}+{}^6C_{m+2}={}^7C_{m+1}+{}^7C_{m+2}={}^8C_{m+2}$. Then ${}^8C_{m+2}>{}^8C_3 \Rightarrow m+2>3$ (or use symmetry) giving $m=2$ as the boundary. Separately solve the permutation ratio for $n$.
${}^6C_m+2{}^6C_{m+1}+{}^6C_{m+2}={}^8C_{m+2}>{}^8C_3\Rightarrow m=2$.
For permutation: $\frac{{}^{n-1}P_3}{{}^nP_4}=\frac{(n-1)(n-2)(n-3)}{n(n-1)(n-2)(n-3)}=\frac{1}{n}=\frac{1}{8}\Rightarrow n=8$.
${}^nP_{m+1}+{}^{n+1}C_m={}^8P_3+{}^9C_2=8\times7\times6+\frac{9\times8}{2}=336+36=372$.
Correct Answer: 4