Question

how many distinct sequences of four letters can be made from the letters in problem if each letter can be used only once and each sequence must begin with l and not end with p?

Answer 1

100

Explanation:

You want the number of distinct sequences of 4 letters from the word "problem" that have "l" as the first letter, and not "p" as the last letter.

Permutations

There are 7 distinct letters in the word "problem". If the first letter of the "words" we're making must be "l", then the remaining three letters of the word are chosen from the remaining 6: "probem".

The number of permutations of 6 things taken 3 at a time is ...

P(6, 3) = 6!/(6 -3)! = 6·5·4 = 120

Exclusions

That count includes the number that end in 'p'. When "l" is the first letter and "p" is the last letter, the middle two letters are chosen from the 5 "robem". There are 5!/(5-2)! = 5·4 = 20 ways to have words starting with "l" and ending in "p". We don't want to count those, so the number of letter sequences we can make is ...

120 -20 = 100

100 distinct sequences of 4 letters can be made with first letter "l" and last letter not "p".