Question

How many different four-letter arrangements can be formed using the six letters A, B, C, D, E and F, if the first letter must be C, one of the other letters must be B, and no letter can be used more than once in the arrangement?

Answer 1

Answer with Step-by-step explanation:

We are given that A,B,C,D,E and F.

We have to find the number of  different four -letter arrangements can be formed using given six letters a,if the first letter must be C and one of the other letters must be B and no letter can be used more than once in the arrangement.

Number of letters=6

We have to arrange  four letter  out of six

After fixing C and B then we choose only two letters out of remaining four letters and repetition is not allowed.

Permutation formula :$nP_r=\frac{n!}{(n-r)!}$

We have n=4 an r=2

Using this formula and substitute the values

Then, we get $4P_2=\frac{4!}{(4-2)!}=\frac{4\times 3\times 2!}{2!}$

$4P_2=4\times 3=12$

Hence, number of different four -letter arrangements can be formed using six letters when repetition is not allowed=12

Answer 2

Answer:

36

Step-by-step explanation:

There is 1 way to make the first letter C and 3 ways to make one of the other letters B. We now have 4 ways to pick the letter for the first remaining spot and 3 ways to pick the letter for the last remaining spot. Thus 1*3*4*3 will get you 36.