Brilliant members
Probability
Level
4
10 Brilliant members are sitting in a row. In how many ways we can select 3 of them such that no two of them are consecutive?
The answer is 56.
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1 solution
Consider we have already selected the 3 members. There will be 4 spaces in total which have to be filled after that i.e. (2 spaces between 2 consecutive selected members) + (2 x 1 space at the extreme) = 4 spaces in total. Now, all we have to do is to find the number of ways in which 7 identical things (identical things as the order of members is already fixed, so we just have to count the no. of seats) can be distributed among 4 different places, provided the spaces in between the consecutive members should not be vacant (at least one member in between). Thus we use distribution of identical things among persons concept of permutations and combinations.
Required no = combination((5+4-1),5) (as 2 members have already been sent to the 2 spaces which can't be vacant)
= 56 ways.