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![[Post New]](/templates/default/images/icon_minipost_new.gif) 7 Jan 2008 13:07:51 IST
Accepted Answer [?]
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yeah...the ans does cum 828.
1) No of ways to select 2 professors:
Case 1: If P1 is selected, P2 can't be selected. We have to select one more frm 3 professors which can be done in 3C1 ways.
The same of we select P2 and not P1.
Hence total no. of ways = 2*3C1 = 2*3=6
Case 2: If both P1 and P2 aren't selected. Then we have to select 2 prof out of 3.
No of ways = 3C2 = 3
Hence total no. of ways to select professors = 6+3=9.
2) The no of ways to select the students:
Case 1: If S2 is present, we hav to select 2 more students out of 9 (S1 can be selctd).
No. of ways = 9C2.
Case 2: If S2 isn't present, we have to select 3 students out of 8 since S1 also can't be selected.
No. of ways = 8C3.
Total no of ways to select students = 9C2+8C3 = 92
Hence total no. of committees possible = 9*92 = 828.
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