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?
+770 The total ways to make sequences of 4 letters using the letters P, R, O, B, L, E, M, using each letter only once, without restrictions, is 7 * 6 * 5 * 4 = 840.
Let's use complementary counting for this problem (So, we find the situations we DON'T want and subtract them from the total)! :D
Ways that a sequence begins with L = 1 * 6 * 5 * 4 = 120
Ways that a sequence ends with P = 6 * 5 * 4 * 1 = 120
Ways that a sequence begins with L and ends with a P = 1 * 5 * 4 * 1 = 20
120 + 120 - 20 = 220
There are 220 ways which we DON'T want!
So, since the total is 840, we subtract 840 from 220 to get \(\fbox{620}\) :D
EDIT: Sorry I misread the problem! This answer is false.
As "CentsLord" calculated, there are 7 nPr 4 =840 permutations.
Each permutation begins with one of the 7 letters this many times =840 / 7 = 120 permutations for each letter. So, letter "L" will begin with 120 permutations. Each of other 6 remaining letters is equally represented in this many ways:120 / 6 =20 ways.
So, 120 permutations beginning with the letter "L" will have 20 of them that end in "P". Since we don't want "P" at the end, then will drop the permutations that end in "P" and we have left:
120 - 20 = 100 permutaions that begin with "L" and NOT end in "P".
