Number of ways in which 10 persons can go in two boats so that there may be 5 on each boat, supposin

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6.Permutation and Combination

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The number of ways in which $10$ persons can go in two boats so that there may be $5 $ on each boat, supposing that two particular persons will not go in the same boat is

A

$\frac{1}{2}{(^{10}}{C_5})$

B

$2{(^8}{C_4})$

C

$\frac{1}{2}{(^8}{C_5})$

D

None of these

Solution

(b) First omit two particular persons, remaining $8 $ persons may be $4$ in each boat. This can be done in $^8{C_4}$ ways.

The two particular persons may be placed in two ways one in each boat. Therefore total number of ways are $ = 2{ \times ^8}{C_4}$.

Standard 11

Mathematics

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(KVPY-2019)