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Solution:

(i) Given: ${ }^{n} C_{7}={ }^{n} C_{5}$

To find : $\mathrm{n}=?$

We know that:

${ }^{n} \mathrm{C}_{r}={ }^{n} \mathrm{C}_{n-r}$

$\Rightarrow{ }^{n} C_{7}={ }^{n} C_{n-7}$

$\Rightarrow{ }^{n} C_{n-7}={ }^{n} C_{5}$

$\Rightarrow n-7=5$

$\Rightarrow \mathrm{n}=7+5=12$

Ans: $\mathrm{n}=12$

(ii) Given: ${ }^{n} C_{14}={ }^{n} C_{16}$

To find: ${ }^{n} C_{28}=?$

We know that

${ }^{n} \mathrm{C}_{r}={ }^{n} \mathrm{C}_{n-r}$

$\Rightarrow{ }^{n} C_{14}={ }^{n} C_{n-14}$

$\Rightarrow{ }^{n} C_{n-14}={ }^{n} C_{16}$

$\Rightarrow n-14=16$

$\Rightarrow n=16+14=30$

$\Rightarrow n=30$

So,

${ }^{\mathrm{n}} \mathrm{C}_{28}={ }^{30} \mathrm{C}_{28}$

$\Rightarrow{ }^{30} \mathrm{C}_{28}=\frac{30 !}{(30-28) ! \times 28 !}$

$\Rightarrow{ }^{30} \mathrm{C}_{28}=\frac{30 !}{2 ! \times 28 !}$

$\Rightarrow{ }^{30} \mathrm{C}_{28}=\frac{30 \times 29 \times 28 !}{2 ! \times 28 !}$

$\Rightarrow{ }^{30} \mathrm{C}_{28}=\frac{30 \times 29}{2 !}$

$\Rightarrow{ }^{30} \mathrm{C}_{28}=\frac{30 \times 29}{2 \times 1}$

$\Rightarrow{ }^{30} \mathrm{C}_{28}=435$

Ans: ${ }^{30} \mathrm{C}_{28}=435$

(iii) Given: ${ }^{n} C_{16}={ }^{n} C_{14}$

To find: ${ }^{n} \mathrm{C}_{27}=?$

We know that:

${ }^{n} C_{r}={ }^{n} C_{n-r}$

$\Rightarrow{ }^{n} C_{14}={ }^{n} C_{n-14}$

$\Rightarrow{ }^{n} C_{n-14}={ }^{n} C_{16}$

$\Rightarrow n-14=16$

$\Rightarrow n=16+14=30$

$\Rightarrow n=30$

So,

${ }^{n} C_{27}={ }^{30} C_{27}$

$\Rightarrow{ }^{30} \mathrm{C}_{27}=\frac{30 !}{(30-27) ! \times 27 !}$

$\Rightarrow{ }^{30} \mathrm{C}_{27}=\frac{30 !}{3 ! \times 27 !}$

$\Rightarrow{ }^{30} \mathrm{C}_{27}=\frac{30 \times 29 \times 28 \times 27 !}{3 ! \times 27 !}$

$\Rightarrow{ }^{30} \mathrm{C}_{27}=\frac{30 \times 29 \times 28}{3 !}$

$\Rightarrow{ }^{30} \mathrm{C}_{27}=\frac{30 \times 29 \times 28}{3 \times 2 \times 1}$

$\Rightarrow{ }^{30} \mathrm{C}_{27}=4060$

Ans: ${ }^{30} \mathrm{C}_{27}=4060$