How many different words can be formed with the letters of 'SKYSCRAPERS' if A and E are together?

Question

a. How many different words can be formed with the letters of 'SKYSCRAPERS' if A and E are together?

b. How many of the words formed from 'SKYSCRAPERS' begin with 'K' and end with 'A'?

Attempt:

I tried solving the a part using $ 9! \cdot 10C2 $ since we can arrange AE among the other 9 words in 10C2 ways but I don't think my attempt is right.

If I am right, I plan to solve the b part using the same method.

This combinatorics question is from my textbook. I came across it when solving different questions. Please assist, if I am wrong with the two parts. Thanks

Answer 1

How many different words can be formed with the letters of SKYSCRAPERS if A and E are together?

The word SKYSCRAPERS has eleven letters. If we place the letters A and E in a box, we have ten objects to arrange. They include $3$ Ss, $1$ K, $1$ Y, $1$ C, $2$ Rs, $1$ P, and $1$ box containing A and E. Choose three positions of the ten positions for the Ss, which can be done in $\binom{10}{3}$ ways, and two of the remaining seven positions for the Rs, which can be done in $\binom{7}{2}$ ways. We are left with five positions in which to arrange five different objects, the four single letters and the box containing A and E, which can be done in $5!$ ways. Finally, arrange A and E within the box, which can be done in $2!$ ways. Hence, the number of arrangements of the letters of SKYSCRAPERS in which A and E are adjacent is $$\binom{10}{3}\binom{7}{2}5!2! = \frac{10!}{7!3!} \cdot \frac{7!}{5!2!} \cdot 5! \cdot 2! = \frac{10!}{3!}$$

How many words formed with the letters of SKYSCRAPERS begin with K and end with A?

As @Bye_World pointed out in the comments, since the positions of K and A are fixed, this amounts to arranging the nine letters of the word SYSCRPERS. Of those nine letters, there are $3$ Ss, $1$ Y, $1$ C, $2$ Rs, $1$ P, and $1$ E. Proceeding as above, we choose three of the nine positions for the Ss, two of the remaining six positions for the Rs, then arrange the four single letters in the remaining four positions, which can be done in $$\binom{9}{3}\binom{6}{2}4! = \frac{9!}{6!3!} \cdot \frac{6!}{4!2!} \cdot 4! = \frac{9!}{3!2!}$$ ways. The factor of $3!$ in the denominator represents the number of ways the three Ss can be permuted among themselves within a given arrangement, and the factor of $2!$ in the denominator represents the number of ways the two Rs can be permuted among themselves within a given arrangement. Since permuting indistinguishable letters among themselves within a given arrangement does not produce an arrangement that is distinguishable from the given arrangement, we would be counting each distinguishable arrangement $3!2! = 6 \cdot 2 = 12$ times if we did not divide by this factor.

Answer 2

You are almost right with part (a).

We have the $9!$ ways to arrange:

    S    K    Y    S    C    R    P    R    S

and with the repetition, this becomes $\dfrac{9!}{3!2!}$.

As the $A$ and $E$ must be together, we cannot place them in $\dbinom{10}{2}$ ways, as not all of these placements mean that $A$ and $E$ are together.

Instead, there are $10$ slots available, and we can place either $AE$ or $EA$ into any of the $10$ slots, giving $20$ ways to place $A$ and $E$ in total.

So, the final answer for part (a) is:

$$\frac{9!}{3!2!}\cdot20=\frac{10!}{3!}$$