+0

# Probability that I get this wrong with out help: 100%

0
166
6

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?

Jul 29, 2020

#1
+533
+5

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

Jul 29, 2020
edited by CentsLord  Jul 31, 2020
#2
0

huh, this actually was wrong.

is there another way to do this or was there maybe some double counting

Wait i think i know what went wrong,

we WANT the sequence to begin with L so idn't think we should have subtracted it from 840.

is that correct?

Guest Jul 29, 2020
edited by Guest  Jul 29, 2020
#3
+1170
-1

We have 1*6*5*3=90 ways to do this

Jul 29, 2020
#4
+1

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".

Jul 29, 2020
#5
+1

Thankyou Guest,

Guest Jul 30, 2020
#6
+533
+5

Sorry! Well, I tried :(

CentsLord  Jul 31, 2020