Question

Consider the word 8 letter word PARALLEL

(a) how many distinguishable ways can you arrange the letters?

(b) In how many distinguishable ways can you arrange the letters so that none of the L's are together?

Answer 1

a)  In the word “parallel”, there are actually 8! permutations of the letters

There are “r” occurs twice

“l” occurs three times,

“r” can be ordered in 2! ways for each permutation

“l” can be similarly ordered in 3! ways.

Therefore, to know that each permutation is unique,

8! must be divided by (2!)(3!).

Since 8! = 4320

= 8!/(2!)(3!) = 3360

b)  total number of arrangements of word PARALLEL: Total letters 8. Repeating letters A and L, the letter A repeating twice and letter L repeating thrice.

The total number of arrangements

8! = 4320

= 8!/(2!)(3!) = 3360

the group of letters that remain together (here L, L, L) are assumed to be a single letter,

all other letters are as usual counted as a single letter. Now find a number of ways as usual

the number of ways of arranging r letters from a group of n letters is equals to ⁿP_r

the final answer is then multiplied by a number of ways we can arrange the letters in that group which has to be stuck together in it (Here L, L, L).

Letters in word PARALLEL: 8 letters

Letters in a new word: LLL, P, A, A, R, E: 6 letters

Total number of word arranging all the letters

= 6! /2! * 3!/3!

total number of arrangements in such a way that all L’s do come together.

= 8!/(2!)(3!) - (6! /2! * 3!/3! )

= 3000