How many $4$-letter words can be formed with letters of the word "SUSTSTE" where the first letter is 'S' and last is 'T'?

Question

I am doing it in this way

S _ _ T

Two letters can be distinguished $= 4c2$

or they can be same, which only possible with "S" $= 1$

$13$. Is it correct?

Answer 1

Your final answer is correct, but your explanation is not.

Since S is in the first position and T is in the last position, we may fill the middle two positions using the letters U, S, T, S, E.

You are correct that both positions may be filled with S's in one way.

Otherwise, we have to use two different letters selected from U, S, T, E. We can fill the second position in the word in $4$ ways and the third position in the word in $3$ ways, giving us $4 \cdot 3 = 12$ possible words where the two middle letters are distinct.

That gives us a total of $1 + 12 = 13$ admissible words, as you found.

However, your explanation is not quite correct. For the case with two different letters, you may select the letters in $\binom{4}{2}$ ways, then must arrange them in the two open slots in $2!$ ways. Notice that $$\binom{4}{2}2! = 12$$ Perhaps you forgot to say that the two different letters can be arranged in the two open positions in $2!$ ways. Otherwise, you made a rather fortunate arithmetic mistake.

Answer 2

Even though first answer is correct. Just explaining it in simple terms.

After removing one S and one T from the letters of SUSTSTE, following letters are remaining

$S = > 2$

$U => 1$

$T=>1$

$E=>1$

With respect to S, there are 4 possibility,

• S $\underline{S}$ $\underline{S}$ T

• S $\underline{S}$ _ T

• S _ $\underline{S}$ T

• S _ _ T

Here are the ways for all 4 possibilities

• First One has only 1 way
• Second one has 3 ways
• Third one has 3 ways
• Fourth one has 3P2 ways(I don't know how to write permutation symbol in latex)

So total ways are $1+3+3+6 = 13$.

Hope it helps!!