Question

If \(\dfrac{{}^{\,n-1}C_{r-1}}{{}^{\,n}C_r}=\dfrac{3}{5}\) and \(\dfrac{{}^{\,n+1}C_{r+1}}{{}^{\,n}C_r}=\dfrac{11}{7}\), then \({}^{\,n}C_{r+3}\div{}^{\,r}C_{n/2}\) is equal to:

• A. \(\frac{3}{5}\)
• B. \(12\)
• C. \(8\)
• D. \(\frac{5}{3}\)

Hint:

Remember the standard identities: \[ \frac{{}^{n-1}C_{r-1}}{{}^{n}C_r}=\frac{r}{n} \] and \[ \frac{{}^{n+1}C_{r+1}}{{}^{n}C_r} =\frac{n+1}{r+1}. \] These are frequently used in advanced combination problems.

Answer 1

The Correct Option is D

Concept: Useful identities of combinations are: \[ \frac{{}^{n-1}C_{r-1}}{{}^{n}C_r}=\frac{r}{n} \] and \[ \frac{{}^{n+1}C_{r+1}}{{}^{n}C_r} =\frac{n+1}{r+1}. \] These identities help us determine the values of \(n\) and \(r\) directly.

Step 1: Use the first given relation.
\[ \frac{{}^{n-1}C_{r-1}}{{}^{n}C_r} =\frac{r}{n} =\frac{3}{5} \] Hence, \[ r=\frac{3n}{5}. \]

Step 2: Use the second relation.
\[ \frac{{}^{n+1}C_{r+1}}{{}^{n}C_r} =\frac{n+1}{r+1} =\frac{11}{7} \] Substituting \(r=\frac{3n}{5}\), \[ \frac{n+1}{\frac{3n}{5}+1} =\frac{11}{7}. \] Cross-multiplying, \[ 7(n+1)=11\left(\frac{3n+5}{5}\right) \] \[ 35n+35=33n+55 \] \[ 2n=20 \] \[ n=10. \] Therefore, \[ r=\frac{3(10)}{5}=6. \]

Step 3: Evaluate the required expression.
\[ {}^{n}C_{r+3} = {}^{10}C_{9} =10 \] and \[ {}^{r}C_{n/2} = {}^{6}C_{5} =6. \] Hence, \[ {}^{10}C_{9}\div{}^{6}C_{5} = \frac{10}{6} = \frac{5}{3}. \] \[ \boxed{\frac{5}{3}} \]