Evaluate the following:

Question:

 

Evaluate the following:

(i) 14C3

(ii) 12C10

(iii) 35C35

(iv) n + 1Cn

(v) $\sum_{r=1}^{5}{ }^{5} C_{r}$

Solution:

(i) We have,    

${ }^{14} C_{3}=\frac{14}{3} \times \frac{13}{2} \times \frac{12}{1} \times{ }^{11} C_{0} \quad\left[\because{ }^{n} C_{r}=\frac{n}{r}{ }^{n-1} C_{r-1}\right]$

$\Rightarrow^{14} C_{3}=364^{2} \quad\left[\because{ }^{n} C_{0}=1\right]$

(ii) We have,

${ }^{12} C_{10}={ }^{12} C_{2} \quad\left[\because{ }^{n} C_{r}={ }^{n} C_{n-r}\right]$

$\Rightarrow{ }^{12} C_{10}={ }^{12} C_{2}=\frac{12}{2} \times \frac{11}{1} \times{ }^{10} C_{0} \quad\left[\because{ }^{n} C_{r}=\frac{n}{r}{ }^{n-1} C_{r-1}\right]$

$\Rightarrow{ }^{12} C_{10}=\frac{12}{2} \times \frac{11}{1} \times 1 \quad\left[\because{ }^{n} C_{0}=1\right]$

$\Rightarrow^{12} C_{10}=66$

(iii) We have,

${ }^{35} C_{35}={ }^{35} C_{0} \quad\left[\because{ }^{n} C_{r}={ }^{n} C_{n-r}\right]$

$\Rightarrow{ }^{35} C_{35}=1 \quad\left[\because{ }^{n} C_{0}=1\right]$

(iv) We have,

${ }^{n+1} C_{n}={ }^{n+1} C_{1} \quad\left[\because{ }^{n} C_{r}={ }^{n} C_{n-r}\right]$

$\Rightarrow{ }^{n+1} C_{n}={ }^{n+1} C_{1}=\frac{n+1}{1} \times{ }^{n} C_{0} \quad\left[\because{ }^{n} C_{r}=\frac{n}{r}{ }^{n-1} C_{r-1}\right]$

$\Rightarrow{ }^{n+1} C_{n}=n+1 \quad\left[\because{ }^{n} C_{0}=1\right]$

(v) We have,

$\sum_{r=1}^{5}{ }^{5} C_{r}={ }^{5} C_{1}+{ }^{5} C_{2}+{ }^{5} C_{3}+{ }^{5} C_{4}+{ }^{5} C_{5}$

$\Rightarrow \sum_{r=1}^{5}{ }^{5} C_{r}={ }^{5} C_{1}+{ }^{5} C_{3}+{ }^{5} C_{3}+{ }^{5} C_{1}+{ }^{5} C_{0} \quad\left[\because{ }^{n} C_{r}={ }^{n} C_{n-r}\right]$

$\Rightarrow \sum_{r=1}^{5} 5_{C_{r}}=2 \times\left(\frac{5}{1} \times 4_{C_{0}}\right)+2 \times\left(\frac{5}{3} \times \frac{4}{2} \times \frac{3}{1} \times 2_{C_{0}}\right)+5_{C_{0}} \quad\left[\because{ }^{n} C_{r}=\frac{n}{r}{ }^{n-1} C_{r-1}\right]$

$\Rightarrow \sum_{r=1}^{5} 5_{C_{r}}=10+20+1=31 . \quad\left[\because{ }^{n} C_{0}=1\right]$

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