Accenture
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Numerical Ability
Probability
10 boys stand in a queue.what is the probability that 5 boys stand between 2 particular boys?
Read Solution (Total 11)
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- 2 boys can be arranged in 4 ways leaving 5 spaces between them
5 boys can be arranged in 8c5 ways
remaining 3 boys can be arranged in 3c3 ways
4*8c5*3c3=672
But total arrangement possible is 8!
So the probabilty is
672/8! Which is equal to 1/60
So ans is 1/60 - 8 years agoHelpfull: Yes(10) No(1)
- 2*8C5*6=672: 2 particular boys are arranged in 2 ways and 5 boys between them in 8C5 ways and remaining 3 are arranged in 3! ways
- 8 years agoHelpfull: Yes(7) No(4)
- 2 boys can be arranged in 4 ways leaving 5 spaces between them
5 boys can be arranged in 8c5 ways
remaining 3 boys can be arranged in 3c3 ways
4*8c5*3c3=672 - 8 years agoHelpfull: Yes(6) No(2)
- no of occurrence is 5!*2!*4!
and total occurrence is 10!
so probability =(5!*2!*4!)/10!=1/630. - 8 years agoHelpfull: Yes(4) No(3)
- let 10 boys stand as
1 2 3 4 5 6 7 8 9 10
if 5 boys are in between two boys
then pairs are:
1-7
2-8
3-9 and
4-10
therefore total no of pairs are 4
total no of stutends=10
therefore P=4/10=.4 - 7 years agoHelpfull: Yes(2) No(0)
- 2*8C5*5!*6 2 particular boys are arranged in 2 ways and 5 boys between them are selected by 8C5 ways and those 5 are arranged in 5! ways and remaining 3 are arranged in 3! ways
- 8 years agoHelpfull: Yes(1) No(1)
- 10! is the answer,i guess
- 8 years agoHelpfull: Yes(0) No(4)
- 2 boys can be arranged in 4 ways leaving 5 spaces between them
5 boys can be arranged in 8c5 ways
remaining 3 boys can be arranged in 3c3 ways
4*8c5*3c3=672 - 8 years agoHelpfull: Yes(0) No(1)
- (8c5*5!*2!*3!)/10!
- 7 years agoHelpfull: Yes(0) No(0)
- (8*8!)/10!
- 7 years agoHelpfull: Yes(0) No(0)
- 2!*5!*3!*4(ways)/10!(total ways)=1/630
ans=1/630.
2!-fixed two boys
5!-between two boys
3!-remaining three boys
4 ways as 1-7,2-8,3-9,4-10
so the ans is 1/630. - 5 years agoHelpfull: Yes(0) No(0)
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