8 Students are arranged in a row. Find the probability that 2 of em, A n B are together..
@albiesriram said:8 Students are arranged in a row. Find the probability that 2 of em, A n B are together..
2.7!/8! = 1/4
Whats the probability when the students are arranged in a circular arrangement.
@amresh_maverick said:Given that f(x) = ax^2 + bx + c and f(4) = 100. If a, b, c are distinct positive integers, then the maximum possible value of a + b + c is a 79 b 87 c 122 d 82 e Data Insufficient
so, whats the answer?
@priyank333333 said:@rahulgaur666 ..approach?
16a+4b +c =100 since a,b,n c are +ve integers
and a,b has higher coefficient as compare to c , therefore to maximise a+b+c put a=1 n b=1 u will get c = 80 hence a+b+c = 82
should be 79 a=1 and b = 2
@ravi6389 That would mean either A or B i.e. P(A+B), while in fact we want both of them to occur simultaneously P(A.B)
@priyank333333 said:@rahulgaur666 ...wat abt distinct positive no.s ? a =1 and b should not be 1 thn ?
sorry didn't notice distinct number, my bad 
if there are two kinds of hats red and blue and at least 5 of each kind in how many ways can the hats be put in each of 5 different boxes?
In How many ways 8 distinct balls can be divided among 3 people such that anyone can get any number of balls?
OA- 3^8
Why cant we not perceive the problem as A+B+C=8
where A,B,C are the number of balls each may receive. In that case number of solution is 8+3-1C2=10C2
Please explain why cant we take this approach and where can we use this approach as well .
@chillfactor @Subhashdec2 @grkkrg @gs4890 PLEASE HELP !
OA- 3^8
Why cant we not perceive the problem as A+B+C=8
where A,B,C are the number of balls each may receive. In that case number of solution is 8+3-1C2=10C2
Please explain why cant we take this approach and where can we use this approach as well .
@chillfactor @Subhashdec2 @grkkrg @gs4890 PLEASE HELP !
@IIM-A2013 said:if there are two kinds of hats red and blue and at least 5 of each kind in how many ways can the hats be put in each of 5 different boxes?
2^5
@Asfakul said:In How many ways 8 distinct balls can be divided among 3 people such that anyone can get any number of balls?
OA- 3^8
Why cant we not perceive the problem as A+B+C=8
where A,B,C are the number of balls each may receive. In that case number of solution is 8+3-1C2=10C2
Please explain why cant we take this approach and where can we use this approach as well .
@chillfactor @Subhashdec2 @grkkrg @gs4890 PLEASE HELP !
u will use multinomial for number of cases like (0,0,8),(1,0,7)....... and so on adding up all these combo u will get 3^8
@rahulgaur666 said:u will use multinomial for number of cases like (0,0,8),(1,0,7)....... and so on adding up all these combo u will get 3^8
Bhai what is a multinomial .can you please shed some light.
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