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# counting question

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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?

Dec 1, 2020

#1
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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.

Dec 1, 2020
#2
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5 nPr 4  should read 6 nPr 4.

Also: 6 nPr 4 - 5 nPr 4 = 240 permutations.

Guest Dec 1, 2020
#3
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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

Dec 1, 2020