Let's go through the questions in the image step by step:

Q1: $nC_4 = 5 \times nC_3$, find $n$

• Use the combination formula $nC_r = \frac{n!}{r!(n-r)!}$.
• We have $nC_4 = \frac{n!}{4!(n-4)!}$ and $nC_3 = \frac{n!}{3!(n-3)!}$.
• Setting $nC_4 = 5 \times nC_3$, solve for $n$.

Q2: $nC_r = 6720$, $nC_{r+1} = 56$; find $n$ and $r$

• Use the relation $nC_{r+1} = \frac{n-r}{r+1} nC_r$ to find $n$ and $r$.

Q3: $18C_r = 18C_{r+2}$; find $C_6$

• Use the identity $nC_r = nC_{n-r}$ to simplify and find $r$.
• Then compute $C_6$.

Q4: $12C_r = 12C_{r+2}$; find $C_3$

• Again, use symmetry in combinations to find $r$, then compute $C_3$.

Q5: $12C_8 = 12C_{r-2}$; find $C_3$

• Use the combination formula and symmetry to find $r$ and then $C_3$.

Would you like detailed solutions for these?