Name: Problem on Properties of Combination
Uploaded: 2019-07-19
Duration: 3 min 49 s
Description: A combination is a selection of items from a collection, such that the order of selection does not matter. Let's learn how to apply an appropriate property o...

Question

A committee of

10

persons is to be formed from a group of

10

women and

8

men. How many possible committees will have at least

5

women? How many possible committees will have men in majority?

Accepted Answer

There are $10$ women and $8$ men.

A committee of $10$ persons is to be formed. If at least $5$ women have been included in a committee, then the possible selections are :

$5$ women and $5$ men, $6$ women and $4$ men,

$7$ women and $3$ men, $8$ women and $2$ men,

$9$ women and $1$ man and all the $10$ women and no man.

$∴$ The number of ways of forming committees such that at least five women are included

$= {}^{10}C_{5} \times^{8} C_{5} + {}^{10}C_{6} \times^{8} C_{4} + {}^{10}C_{7} \times {}^{8}C_{3} + {}^{10}C_{8} \times {}^{8}C_{2} + {}^{10}C_{9} \times^{8} C_{1} + {}^{10}C_{10} \times^{8} C_{0}$

$8$

$= (\frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} \times \frac{8 \times 7 \times 6}{1 \times 2 \times 3}) + (\frac{10 \times 9 \times 8 \times 7}{1 \times 2 \times 3 \times 4} \times \frac{8 \times 7 \times 6 \times 5}{1 \times 2 \times 3 \times 4}) + (\frac{10 \times 9 \times 8}{1 \times 2 \times 3} \times \frac{8 \times 7 \times 6}{1 \times 2 \times 3}) + (\frac{10 \times 9}{1 \times 2} \times \frac{8 \times 7}{1 \times 2}) + (10 \times 8) + (1 \times 1)$

$= (252 \times 56) + (210 \times 70) + (120 \times 56) + (45 \times 28) + 80 + 1$

$= 14112 + 14700 + 6720 + 1260 + 80 + 1$

$= 36873$ .

Men in majority :

In a committee of $10$ persons if men are in majority, then they are $6$ or $7$ or $8$ in numbers.

The number of committees in which men are in majority

$= {}^{8}C_{6} \times {}^{10}C_{4} + {}^{8}C_{7} \times {}^{10}C_{3} + {}^{8}C_{8} \times {}^{10}C_{2}$

$= \frac{8!}{6! 2!} \times \frac{10!}{4! 6!} + \frac{8!}{7! 1!} \times \frac{10!}{3! 7!} + \frac{8!}{8! 0!} \times \frac{10!}{2! 8!}$

$= (\frac{8 \times 7}{1 \times 2} \times \frac{10 \times 9 \times 8 \times 7}{1 \times 2 \times 3 \times 4}) + (8 \times \frac{10 \times 9 \times 8}{1 \times 2 \times 3}) + (1 \times \frac{10 \times 9}{1 \times 2})$

$= (28 \times 210) + (8 \times 120) + 45$

$= 5880 + 960 + 45$

$= 6885$ .

A committee of $10$ persons is to be formed from a group of $10$ women and $8$ men. How many possible committees will have at least $5$ women? How many possible committees will have men in majority?

Important Questions on Permutations and Combinations

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