Question

Counting Arrangements A password is going to be formed by rearranging all of the letters of the word WILLAMETTE. (i). How many total different arrangements are possible? (ii). How many if the two L's must be next to each other (LL)? (iii). How many if W cannot be first and E cannot be last? (So LIWMEELATT is okay, but LIWMELATTE is not.)

Answer 1

Word is WILLAMETTE.

1)

Total different arrangements possible = (10!)/((2!)(2!)(2!)) = 3628800/8 = 453600

Total letters = 10

( 2L , 2T , 2E , W , I , A , M )

2)

two L's must be next to each other.

( 1W , 2T , 2E , 1(LL) , 1A , 1M , 1I)

No. of arrangements = (9!)/((2!)(2!)) = 362880/4 = 90720

3)

if W cannot be first and E cannot be last.

Equivalent to

Total - W can be first or E can be last.

Arrangements = Total - [ W is first ( fixed ) + E is last ( fixed ) - W is first and E is last ( both fixed ) ]

= 453600 - [ (9!)/((2!)(2!)(2!)) + (9!)/((2!)(2!)) - (8!)/((2!)(2!)) ]

= 453600 - [ 45360 + 90720 - 10080 ]

= 327600

Hope this helps.

Thanks.

Comment if you have any questions.