Question

If $n$ be the number of $4$ element subset that can be chosen from the set of first $12$ natural number so that at least $2$ of the four numbers are consecutive , then the last digit of $n$ is

Answer 1

Total number of subsets $=$ ${}^{12}{C}_4=495$
Now, find the subsets where the selected 4 numbers are non-consecutive.
For each such case, the non-consecutive numbers occupy any four of the gaps created by remaining 8 natural numbers.
Total number of gaps formed by 8 numbers is 9.
So the number of solutions is equivalent to filling any 4 out of 9 gaps obtained. This can be done in ${}^9{C}_4=126$ ways.
Hence, the number of subsets where atleast two numbers are consecutive is $495-126=369$