In a school there are 8 teachers including headmaster. In how many ways can a committee of 5 is to be formed without the headmaster
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Answered by
2
it is formed by 8C5 ways
8
C. = 8*7*6/3*2*1=56
5
the answer is 56 ways
8
C. = 8*7*6/3*2*1=56
5
the answer is 56 ways
Answered by
0
Answer:
The general formula which is to be used here is: (x!)/[(y!){(x-y)!}]
In the above question, x=8 and y=5
=> (8!)/[(5!){(8-5)!}]
=> (8*7*6*5*4*3*2*1)/[(5*4*3*2*1)(3)!]
=> (8*7*6)/(3*2*1)
=> 8*7
=> 56
Therefore, there are 56 combinations in which committees of 5 people from 8 people can be formed.
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