Question

Show your work, please

2. 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

(i) Total different arrange ments possible 10! = 453600 2!*2!*a! If the word WILLAMETTE is excluded to make the password, then the total possible arrangements = 453599 (ii) If the two ts must be next to each then the possible arrangements, other, 11 181440 2 x 91 2! *2!

(ii) W cannot be first and cannot be last. le Words starting with W is not possible and words ending with E is not possible. Number of words starting with W giren by, is 9!. 2! x2! 2! 45360 (A) Number of words ending with L 2x 91 90720 =(B) 2!*2! xd! Number of words begining with W and ending with L 2x 8 ! 10080 2!*2! X2! - (6) Therefore number of words to which does not start with and does not end with L = (A + B)-c = 126000

(i) Total different arrangements possible= n! / (p!*q!*r!)

where n=10 (total number of letters present in the word)

p=2 (number of repetitions of L)

q=2 (number of frepetirions of T)

r=2 (number of repetitions of E)

Answer=453600

(ii) Answer=181440

(iii) W cannot be first and L cannot be last. i.e. Words starting with W is not possible and words ending with E is not possible.

Number of words starting with W = (A)

Number of words ending with L =(B)

Number of words beginning with W and ending with L =(C)

Therefore number of words to which does not start with W and does not end with L

= (A + B)-C = 126000