MHT CET · Maths · Binomial Theorem
If \({ }^{11} \mathrm{C}_4+{ }^{11} \mathrm{C}_5+{ }^{12} \mathrm{C}_6+{ }^{13} \mathrm{C}_7={ }^{14} \mathrm{C}_{\mathrm{r}}\), then value of \(\mathrm{r}\) is
- A 11
- B 14
- C 7
- D 3
Answer & Solution
Correct Answer
(C) 7
Step-by-step Solution
Detailed explanation
\(
{ }^{11} \mathrm{C}_4+{ }^{11} \mathrm{C}_5+{ }^{12} \mathrm{C}_6+{ }^{13} \mathrm{C}_7={ }^{14} \mathrm{C}_{\mathrm{r}}
\)
We know that \({ }^n C_r+{ }^n C_{r-1}={ }^{n+1} C_r\)
\(
\begin{aligned}
& \therefore{ }^{14} \mathrm{C}_{\mathrm{r}}=\left({ }^{11} \mathrm{C}_4+{ }^{11} \mathrm{C}_5\right)+{ }^{12} \mathrm{C}_6+{ }^{13} \mathrm{C}_7 \\
& =\left({ }^{12} \mathrm{C}_5+{ }^{12} \mathrm{C}_6\right)+{ }^{13} \mathrm{C}_7 \\
& ={ }^{13} \mathrm{C}_6+{ }^{13} \mathrm{C}_7={ }^{14} \mathrm{C}_7 \\
& \therefore \mathrm{r}=7
\end{aligned}
\)
{ }^{11} \mathrm{C}_4+{ }^{11} \mathrm{C}_5+{ }^{12} \mathrm{C}_6+{ }^{13} \mathrm{C}_7={ }^{14} \mathrm{C}_{\mathrm{r}}
\)
We know that \({ }^n C_r+{ }^n C_{r-1}={ }^{n+1} C_r\)
\(
\begin{aligned}
& \therefore{ }^{14} \mathrm{C}_{\mathrm{r}}=\left({ }^{11} \mathrm{C}_4+{ }^{11} \mathrm{C}_5\right)+{ }^{12} \mathrm{C}_6+{ }^{13} \mathrm{C}_7 \\
& =\left({ }^{12} \mathrm{C}_5+{ }^{12} \mathrm{C}_6\right)+{ }^{13} \mathrm{C}_7 \\
& ={ }^{13} \mathrm{C}_6+{ }^{13} \mathrm{C}_7={ }^{14} \mathrm{C}_7 \\
& \therefore \mathrm{r}=7
\end{aligned}
\)
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