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A committee of 7 peoples has to be formed from 8 men and 4 women. In how many ways can this be done when the committee consists of at most 3 women?

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#### Solution

Number of men = 8

Number of women = 4

Number of peoples in the committee = 7

At most 3 women

The 7 members must contain at most 3 women,

∴ We have the following possibilities

**(i)** No women + 7 men

**(ii)** 1 women + 6 men

**(iii)** 2 women + men

**(iv)** 3 women + 4 men

**Case (i):** 0 women + 7 men

The number of ways of selecting 0 women from 4 women is = ^{4}C_{0}

The number of ways of selecting 7 men from 8 men is = ^{8}C_{7}

Total number of ways = ^{4}C_{0} × ^{8}C_{7}

**Case (ii):** 1 women + 6 men

The number of ways of selecting 1 woman from 4 women is = ^{4}C_{1}

The number of ways of selecting 6 men from 8 men is = ^{8}C_{6}

Total number of ways = ^{4}C_{1} × ^{8}C_{6}

**Case (iii):** 2 women + 5 men

The number of ways of selecting 2 women from 4 women is = ^{4}C_{3}

The number of ways of selecting 4 men from 8 men is = ^{8}C_{4}

∴ Total number of ways = ^{4}C_{3} × ^{8}C_{4}

∴ The required number of ways of forming the committee

= ^{4}C_{0} × ^{8}C_{7 }× ^{4}C_{1} × ^{8}C_{6} + ^{4}C_{2} × ^{8}C_{5 }+ ^{4}C_{3} × ^{8}C_{6}

= `1 xx (8!)/(7!(8 - 7)!) + (4!)/(1!(4 - 1)!) xx (8!)/(6!(8 - 6)!) + (4!)/(2!(4 - 2)!) xx (8!)/(5!(8 - 5)!) + (4!)/(3!(4 - 3)!) xx (8!)/(4!(8 - 4)!)`

= `(8!)/(7 xx 1!) + (4!)/(11 xx 3!) xx (8!)/(6! xx 2!) + (4!)/(2! xx 2!) xx (8!)/(5! xx 3!) + (4!)/(3! xx 1!) xx (8!)/(4! xx 4!)`

= `(8 xx 7!)/(7!) + ( xx 3!)/(3!) xx (8 xx 7 xx 6!)/(6! xx 2!) + (4 xx 3 xx 2!)/(2! xx 2!) xx (8 xx 7 xx 6 xx 5!)/(5! xx 3!) + (4 xx 3!)/(3!) xx (8 xx 7 xx 6 xx 5 xx 4!)/(4! xx 4!)`

= `8 + 4 xx (8 xx 7)/(2 xx 1) + (4 xx 3)/(2 xx 1) xx (8 xx 7 xx 6)/(3 xx 2 xx 1) + 4 xx (8 xx 7 xx 6 xx 5)/(4 xx 3 xx 2 xx 1)`

= 8 + 4 × 4 × 7 + 2 × 3 × 8 × 7 + 4 × 2 × 7 × 5

= 8 + 112 + 336 + 280

= 736

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