Question

The following parts refer to the letters: LAMEFIREALARM. Recall that "word" means distinguishable letter arrangements. (a) How many words can be made with the above letters such that the M's are not next to each other? (b) What is the probability that, given a random word made from the letters above, the M's will be separated by at legist two letters?

Answer 1

Answer:

Step-by-step explanation:

The word "LAMEFIREALARM" word consists of

2L , 2M, 2E , 2R , 3A , F and I

no of words with the above letters such that M's are not together

=Total No of words- M's are together

Total no of words$=\frac{13!}{4\times 2!\times 3!}=64,864,800$

When M's are together . considering 2 M's as one

therefore there are $\frac{12!}{3\times 2!\times 3!}=9,979,200$

No of ways$=\frac{13!}{4\times 2!\times 3!}-\frac{12!}{3\times 2!\times 3!}=54,885,600$

(b)No of ways such that M's are separated by at least 2 letters

at least  2 letter means 2 letter, 3 letter ......11 letters

So we have to subtract no of ways where there are 1 letter in between M's from total no of ways where 2 M's are not next to each other

No of ways in which there is 1 letter between 2 M's

This can be done by considering 11 cases

In first case Place first M in Starting Position and 2 M on third place

Second case place First M in 2 nd Position and second M on 4 th place

Similarly For 11th case

Place first M in 11th place and second M on 13th place

Total ways

$=\frac{11\times 11!}{3!\times 2!\times 2!\times 2!}$

So , the total number of arrangements of the letters of the word LAMEFIREALARM where the two M's are separated by atleast two letters is

$=\frac{13!}{4\times 2!\times 3!}-\frac{12!}{3\times 2!\times 3!}-\frac{11\times 11!}{3!\times 2!\times 2!\times 2!}$

$=45738000$