What is the probability that when the letters of the word SERENDIPITY

My strategy was first to calculate the probability to have T and E

p(T) is 1/11 and P(E) 2/11

Then my second step was to calculate the P to have both T and E as the first letter, 1/11 + 2/11 = 3/11

To find the final value I have subtract 1 - 3/11 = 8/11

Can anyone confirm if my strategy was correct or not?

Solution:

We use the indistinguishable permutations formula to calculate the total number of ways to arrange the letters in the word SERENDIPITY: 11! / (2! x 2!). (Note that the two E’s and two I’s are indistinguishable, so we must divide 11! by 2! twice to account for this.)

If T is the first letter, then there are 10! / (2! x 2!) ways to arrange the remaining letters.

If (one of the) E is the first letter, then there are 10! / 2! ways to arrange the remaining letters.

We see that the total number of ways to arrange the letters, taking into account the stated restrictions, is:

[11! / (2! x 2!) - 10! / (2! x 2!) - 10! / 2!].

Thus, the probability of this occurrence is:

[11! / (2! x 2!) - 10! / (2! x 2!) - 10! / 2!] / [11!/(2! x 2!)]

= 1 - [10!/(2! x 2!)]/[11!/(2! x 2!)] - [10!/2!]/[11!/(2! x 2!)]

= 1 - 10!//11! - [10! x 2!/(2! x 2!)]/[11!/(2! x 2!)]

= 1 - 1//11 - [10! x 2!/]/11!

= 1 - 1/11 - [1 x 2]/11

= 1 - 3/11 = 8/11

Answer: C

Can someone please explain the part where - Number of permutations with E as first alphabet = 10!2! ??

=> 1 - prob( 1st letter is T or E)
1 - (1/11+2/11)
1-3/11
8/11