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A committee of 7 has to be formed out of 9 boys and 4 girls. In how many ways can be this be done when the committee consist of 

(i) exactly 3 girls 

(ii) At most 3 girls?

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A committee of 7 has to be formed from 9 B and 4 G. 

(i) exactly 3 girls

\(=^9C_4\times^4C_3\)

\(\frac{9!}{4!\space5!} \times \frac{4!}{3!\space1!}=\frac{9\times8\times7\times6\times5!}{3\times2\times5!}\)

\(=72\times7 = 504\)

(ii) at most 3 girls 

(a) No girl and 7 boys 

(b) 1 girl and 6 boys 

(c) 2 girls and 5 boys 

(d) 3 girls and 4 boys 

∴ The committee consists of at most 3 girls

\(=^4C_0\times ^9C_7+^4C_1\times^9C_6+^4C_2\times^9C_5+^4C_3\times^9C_4\)

= 36 + 336 + 1296 + 504 

= 2172

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