Question #283540

A committee of 8 is to be formed from 16 men and 10 women.





In how many ways can the committee be formed if





i) there are no restrictions.





ii) there must be 4 men and 4 women.





iii) there should be an even number of women.





iv) there are more women than men.





v) there are atleast 6 men.

1
Expert's answer
2021-12-30T01:59:54-0500

Solution:

Given, 16 men and 10 women.

i) No. of ways when there are no restrictions

= 26C8=1562275=\ ^{26}C_8 \\=1562275


ii) No. of ways when there must be 4 men and 4 women

= 16C4× 10C4=382200=\ ^{16}C_4\times \ ^{10}C_4 \\=382200

iii) No. of ways when there should be an even number of women

= 16C6× 10C2+ 16C4× 10C4+ 16C2× 10C6+ 16C0× 10C8=8008×45+1820×210+120×210+1×45=767805=\ ^{16}C_6\times \ ^{10}C_2+\ ^{16}C_4\times \ ^{10}C_4+\ ^{16}C_2\times \ ^{10}C_6+\ ^{16}C_0\times \ ^{10}C_8 \\=8008\times45+1820\times210+120\times210+1\times45 \\=767805

iv) No. of ways when there are more women than men

= 16C3× 10C5+ 16C2× 10C6+ 16C1× 10C7+ 16C0× 10C8=560×252+120×210+16×120+1×45=168285=\ ^{16}C_3\times \ ^{10}C_5+\ ^{16}C_2\times \ ^{10}C_6+\ ^{16}C_1\times \ ^{10}C_7+\ ^{16}C_0\times \ ^{10}C_8 \\=560\times 252+120\times210+16\times120+1\times45 \\=168285

v) No. of ways when there are at least 6 men

= 16C6× 10C2+ 16C7× 10C1+ 16C8× 10C0=8008×45+11440×10+12870×1=487630=\ ^{16}C_6\times \ ^{10}C_2+\ ^{16}C_7\times \ ^{10}C_1+\ ^{16}C_8\times \ ^{10}C_0 \\=8008\times45+11440\times10+12870\times1 \\=487630


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