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abs4

  • 3 years ago

A committee of 5 people is chosen from 6 men and 4 women. in how many ways can this be done 1) if there must be more men than women on the commitee. 2) if there must be 3 men and 2 women, and one particular woman refuses to be on the commitee with one particular man

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  1. nitz
    • 3 years ago
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    Permutations and combinations....

  2. sauravshakya
    • 3 years ago
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    More men is when No. of men is 3 or No. of men is 4 or No. of men is 5

  3. sauravshakya
    • 3 years ago
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    Which means there must be atleast three men

  4. sauravshakya
    • 3 years ago
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    6C3*7C2 First we choose at least three 3 men then choose any two people from the remaining people

  5. sauravshakya
    • 3 years ago
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    got it?

  6. abs4
    • 3 years ago
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    the answer is 186

  7. abs4
    • 3 years ago
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    the marking scheme says choose 4 men and 1 women or 5 men and no women. i don't understand why we can't choose 3 men and 2 women.

  8. sauravshakya
    • 3 years ago
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    u can do it too

  9. sauravshakya
    • 3 years ago
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    I mean u can choose 3 men and 2 women

  10. abs4
    • 3 years ago
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    yeah but i wonder why in my mark scheme it's not given. that's why the total answer is 186 but i got 180. by choosing 4 men, 1 women and 3 men 2 women. thanks anyways..:)

  11. abs4
    • 3 years ago
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    do you know how to do the second part?

  12. sauravshakya
    • 3 years ago
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    6C3 * 4C2 + 6C4 *4C1 +6C5 =120+60+6=186

  13. sauravshakya
    • 3 years ago
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    U forgot the case when all are men

  14. sauravshakya
    • 3 years ago
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    Can someone tell me why 6C3 *7C2 doest work

  15. shubhamsrg
    • 3 years ago
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    I was also wondering the same thing. hmm..

  16. shubhamsrg
    • 3 years ago
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    M1 M2 M3 M4 M5 M6 W1 W2 W3 W4 Can we try to rationalize ?

  17. saloniiigupta95
    • 3 years ago
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    @sauravshakya, can you explain how u got 6C3*7C2 ?

  18. sauravshakya
    • 3 years ago
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    First we choose at least three 3 men then choose any two people from the remaining people

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