A commitee of 5 from 6 gents and 4 ladies with utmost 2 ladies? My answer: $$\binom{4}{0}\binom{6}{5}+ \binom{4}{1}\binom{6}{4}+\binom{4}{2}\binom{6}{3}=246$$ ways. Is it correct?
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Please demonstrate an effort to solve the problem. What have you tried? Where are you stuck at? What can't you understand? – Gummy bears Sep 11 '15 at 16:35
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My answer.4c06c5+4c16c4+4c2*6c3= 246 ways. Is it correct? – Ishaque Sep 11 '15 at 16:36
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1Utmost two ladies refers to either 0 ladies, 1 lady, or 2 lady. Make separate cases and then add them! – Gummy bears Sep 11 '15 at 16:37
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The strategy you used to calculate the answer is correct. However, $246$ is incorrect. Check your arithmetic. – N. F. Taussig Sep 11 '15 at 16:44
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1Thanks..arithmetic mistake.... 186 – Ishaque Sep 11 '15 at 16:46
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Wolfram Solution – Dr Xorile Sep 11 '15 at 16:48
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And why at most two women. That's sexist! – Dr Xorile Sep 11 '15 at 16:49
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Thats some where about scientists – Ishaque Sep 11 '15 at 16:51