reto wrote:

What is the probability of randomly selecting an arrangement of the letters of the word "MEDITERRANEAN" in which the first letter is E and the last letter is R?

A. 1/13

B. 1/20

C. 1/26

D. 1/50

E. 1/100

Is this a tough combinations/probability question? I have no difficulty level. I find it hard.

Dont know about the difficulty but if you adopt 1 method to deal with arrangements, you can solve all the questions once you realise what is getting asked.

We have the word mediterranean which has 13 letters out of which E is 3, R/A/N are 2 each.

Again, probability = favorable outcomes / total outcomes

Total outcomes = arrangement of 13 letters with 3 Es and 2 each of R/A/N = 13! / (3!*2!*2!*2!)

Favorable outcomes = E <<11 letters>> R. Now E can be selected in 1 way out of 3 Es as all are same . Similarly R can be selected in 1 way out of 3 Rs as all Rs are same.

Thus we have : 1 (for E) <<11 letters with 2 Es , 2 A/N remaining while rest of them are 1 each>> 1 = 1* 11!/(2!*2!*2!) * 1

Dividing Favorable outcomes by Total outcomes , we get ----> \(\frac{1* 11!/(2!*2!*2!) * 1}{13! / (3!*2!*2!*2!)}\) = 1/26. Thus B is the correct answer.