How many distinct sequences of four letters can be made from the letters in EQUALS if each sequence cannot begin with L and cannot end with Q, and no letter can appear in a sequence more than once?

Guest Dec 1, 2020

#1**0 **

E Q U A L S

5 nPr 4 = 360 permutations.

Each 4-letter permutation begins with one of the 6 letters this many times: 360 / 6=60 times, and that many times with the letter L, and ends with the remaining 5 letters this many times:60 / 5 = 12 permutations ending in the letter Q.

Therefore, there are: 360 - 60 - 60=**240 permutations with no L at the beginning and no Q at the end.**

Guest Dec 1, 2020

#3**+1 **

I interpreted this question a little differently

"* Each* sequence cannot begin with L and cannot end with Q "

P(6,4) = 360 possibile sequences

Subtract out the ones that begin with an L and end with a Q

L _ _ Q

The other 2 letters can be chosen in C(4,2) ways = 6 ways and each of these can be arranged in 2 ways =

6 * 2 = 12 words beginning with L and ending with Q

So

360 - 12 =

348 sequences

CPhill Dec 1, 2020