Q:

How many ways can eight letters be arranged into groups of five where order matters and the first two letters are already chosen?

Accepted Solution

A:
Answer:120Step-by-step explanation:Since we're dealing with a problem where the order matters and the first two letters are already chosen we need to subtract the number of letters and the number of available slots per group.We use the permutation formula to find the answer, but before that let's check values.n = 8k = 5Now since there are two letters already chosen we have to deduct two from both the value of n and k.n = 6k = 3Now we can use the permutation formula:[tex]_{n}P_{k}=\dfrac{n!}{(n-k)!}[/tex][tex]_{6}P_{3}=\dfrac{6!}{6-3)!}[/tex][tex]_{6}P_{3}=\dfrac{6!}{3!}[/tex][tex]_{6}P_{3}=\dfrac{6*5*4*3*2*1}{3*2*1}[/tex]The 3*2*1 cancels out and leaves us with:[tex]_{6}P_{3}=6*5*4[/tex][tex]_{6}P_{3}=120[/tex]So there are 120 possible ways to arrange eight letters into groups of five where order matters and the first two letters are already chosen.