# Examples

**Example**

Evaluate

^{10}C_{4}^{12}C_{7}^{6}C_{6}

**Solution**

^{10}C_{4}=^{12}C_{7 }=^{6}C_{6}=

**Example**

- If
^{2n}C_{r }=^{2n}C_{r+2}, find r.

**Solution**

- We shall use the result
^{n}C_{r }=^{n}C_{s}â‡’ r = s or r + s = n

^{2n}C_{r }=^{2n}C_{r+2}

Thus we can conclude as follows

â‡’ r = r + 2 or r + (r + 2) = 2n

But r â‰ r + 2, therefore 2r + 2 = 2n â‡’ r = n - 1.

- If
^{n}C_{10}=^{n}C_{14}find^{n}C_{20}and^{25}C_{n. }

**Solution**

^{n}C_{10}=^{n}C_{14}

Thus

^{n}C

_{20}=

^{24}C

_{20}=

= 23 Ã— 22 Ã— 21 = 10626

and

^{25}C

_{n}=

^{25}C

_{24}=

^{25}C

_{1}= 25.â€‹

- â€‹If
^{18}C_{r}=^{18}C_{r+2}, find^{r}C_{5 }

**Solution**

^{18}C_{r}=^{18}C_{r+2}â‡’ r + r + 2 = 18 or r = 8.

^{r}C

_{5}=

^{8}C

_{5}=

- If
^{n+2}C_{8}:^{n-2}P_{4 }= 57 : 16 find n.

**Solution**

^{n+2}C_{8}:^{n-2}P_{4}= 57 : 16

â‡’ (n+2) (n+1) n (n-1) = (57) (2520)

â‡’ (n

^{2}+ n - 2) (n

^{2}+ n) = 14360 (1)

Put n

^{2}+ n - 1 = y, so that (1) can be written as (y - 1) (y + 1) = 143640

â‡’ y

^{2}- 1 = 143640 â‡’ y

^{2}= 143641 = 379

^{2}â‡’ y = 379 [Q y > 0]

â‡’ n

^{2}+ n - 1 = 379 â‡’ n

^{2}+ n - 380 = 0 [putting the value of y]

â‡’ .

As n cannot be negative, n = 19.

- If
^{n}P_{r}=^{n}P_{r+1}and^{n}C_{r }=^{n}C_{r-1 }; find n and r.

**Solution**

- We are given
^{n}P_{r}=^{n}P_{r+1 }

Next,

^{n}C

_{r}=

^{n}C

_{r-1 }[given]

â‡’ n - r + 1 = r

â‡’ n - 2r = -1 (2)

From (1) and (2), n = 3, r = 2.