Given that,
2N<N!
N is the sum of numbers of two dice.
⇒2≤N≤12
Let us check the given condition 2N<N!
N=1 (not possible) →0
N=2 (not possible) →1
N=3 (not possible) →2
N=4 (possible)
We need to find the probability for N≥4.
⇒P(N≥4)=1−P(N=2)−P(N=3)
=1−361−362
=1211=nm
Now let us find 4m−3n.
=4×11−3×12=8.
Hence, the required answer is 8.