anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Permutations and combinations....
anonymous
  • anonymous
More men is when No. of men is 3 or No. of men is 4 or No. of men is 5
anonymous
  • anonymous
Which means there must be atleast three men

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More answers

anonymous
  • anonymous
6C3*7C2 First we choose at least three 3 men then choose any two people from the remaining people
anonymous
  • anonymous
got it?
anonymous
  • anonymous
the answer is 186
anonymous
  • anonymous
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.
anonymous
  • anonymous
u can do it too
anonymous
  • anonymous
I mean u can choose 3 men and 2 women
anonymous
  • anonymous
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..:)
anonymous
  • anonymous
do you know how to do the second part?
anonymous
  • anonymous
6C3 * 4C2 + 6C4 *4C1 +6C5 =120+60+6=186
anonymous
  • anonymous
U forgot the case when all are men
anonymous
  • anonymous
Can someone tell me why 6C3 *7C2 doest work
shubhamsrg
  • shubhamsrg
I was also wondering the same thing. hmm..
shubhamsrg
  • shubhamsrg
M1 M2 M3 M4 M5 M6 W1 W2 W3 W4 Can we try to rationalize ?
anonymous
  • anonymous
@sauravshakya, can you explain how u got 6C3*7C2 ?
anonymous
  • anonymous
First we choose at least three 3 men then choose any two people from the remaining people

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