IIT JEE , AIEEE Forum - Mathematics , Algebra - Permutation & combinations ( ARIHANT ALGEBRA ) page no 426 <q no 8>

empty_brains09

There are 5 professors (P1,P2,P3,P4,P5) and 10 students (S1,S2,.......S10) out of whom a committee of 2 professors and 3 students is to be formed. Find the number of committees in which P1 and P2 cannot serve together and S1 will serve if S2 is present in the committee ..

Q.8 / Arihant ALGEBRA / LEVEL III / Page no. 426

The answer given is 48 but I got 828.. Please chek this out and reply as soon as possible ...

panks_01

there r 5 prof out of which 2 r to b selected.........so d no. of combns wil b

⁵C₂= 5! / (3!*2!) = 10

but dis includes d combn of P1 n P2 also wich shud not b not included

so d prof. can b selected in 10 - 1 = 9 wez

nw 3 students out of 10 r to b selected

consider 2 cases:

i) S2 is absent

then S1 wil also b absent

so all 3 students wil b seleced frm d remaining 8

so ⁸C₃ = 8! / (3!*5!) = 56 wez

ii) S2 is present

then S1 may or may not b present

let us consider dat S1 is present

then d third student wil b selected frm d remaining 8 = 8 wez

if S1 is not present

then d other 2 students can b selected frm d remaining 8

so ⁸C₂= 8! / (6!*2!) = 28 wez

i think d method is dis........but not getting d ans as 48

wat method did u use?

rhd92781

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.

krishna.gopal

828 is right answer for this question. I think there is some misprinting in the answer

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