Answer to Question #350641 in Discrete Mathematics for Kholeen

Question

Patrick has assignments in 5 subjects he can only do two assignments in how many ways can he do two assignments

Answer 1

We use the formula of combinations $C_k^n=\frac{n!}{k!(n-k)!}$, $n>k$.

We have $5$ subjects, so $n=5$.

Patrick can only do two assigments, so $k=2$.

Number of ways can do two assignments equals $C_2^5=\frac{5!}{2!(5-2)!}=\frac{5!}{2!3!}=\frac{5\cdot4\cdot3\cdot2\cdot1}{2\cdot1\cdot3\cdot2\cdot1}=5\cdot2=10$.