Question

If the letters of the word "FATHER" are arranged at random, what is the
chance that the two letters A and R will be at the either extremes ?​

Answer 1

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Answer 2

Given:

The given word is FATHER

To find:

We have to find the probability that the words arranged will have A and R at either extrimities

Solution:

Total number of words in FATHER = 6

The total number of arrangements made at random = 6!

Let us arrange A and R at exrimities in 2! ways as follows

A _ _ _ _ R  and  R _ _ _ _ A

Now the remaining letters F T H E can be arranged in 4! ways

∴Arrangements made with A and R at either extremes = 4!2!

Probability = $\frac{4!2!}{6!}$ = $\frac{48}{720}$ = $\frac{1}{15}$ = 0.067

∴The probability of arranging the letters of the word FATHER with A and R at either extremes = 0.067