Permutations and Combinations: Question 9

Worked solution

Part (a): Total arrangements of the 9 letters

The word ADDRESSES has $9$ letters: A, D, D, R, E, S, S, E, S. Reading off the letter counts: D appears $2$ times, E appears $2$ times, S appears $3$ times, and A, R each appear once.

If all $9$ letters were different, there would be $9!$ arrangements. Since swapping identical letters with each other doesn’t create a new arrangement, divide by the factorial of each repeated count:

$\frac{9!}{2!\,2!\,3!} = \frac{362880}{2\times2\times6} = \frac{362880}{24} = 15120.$

Check (positions method): choose which $3$ of the $9$ positions hold an S: $\binom{9}{3}=84$. From the remaining $6$ positions, choose which $2$ hold a D: $\binom{6}{2}=15$. From the remaining $4$ positions, choose which $2$ hold an E: $\binom{4}{2}=6$. The last $2$ positions take the distinct letters A, R in either of $2!=2$ orders. $84\times15\times6\times2 = 15120.$ Both methods agree: $15120$.

Part (b): The three Ss together (block method)

Glue the three Ss into a single block, SSS. Because the three Ss are identical, there is no extra factor for “ordering inside the block”, any rearrangement of the Ss inside it looks the same.

This leaves $7$ objects to arrange: the block SSS, together with A, D, D, R, E, E (that’s $6$ single letters $+$ $1$ block $=7$ objects). Among these, D still repeats twice and E still repeats twice:

$\frac{7!}{2!\,2!} = \frac{5040}{4} = 1260.$

Check (proportion argument): among all $\binom{9}{3}=84$ ways to choose the $3$ (identical) positions for the three Ss out of $9$ slots, the number of ways that give $3$ consecutive positions is $7$ (the runs $(1,2,3),(2,3,4),\ldots,(7,8,9)$). So the fraction of arrangements with the Ss all together should be $\frac{7}{84}=\frac{1}{12}$. Checking against part (a)‘s total: $15120\times\frac{1}{12} = 1260.$ Both methods agree: $1260$.

Part (c): Arrangements starting with A

Fix A in the first position. This leaves the other $8$ letters (D, D, R, E, S, S, E, S) to arrange in the remaining $8$ positions, where D still repeats twice, E still repeats twice, and S still repeats three times:

$\frac{8!}{2!\,2!\,3!} = \frac{40320}{24} = 1680.$

Check (proportion argument): A occupies exactly $1$ of the $9$ letter-slots overall, and since A is not repeated, by symmetry A is equally likely to sit in any of the $9$ positions across all arrangements from part (a). So the fraction of arrangements with A first is $\frac{1}{9}$: $15120\times\frac{1}{9} = 1680.$ Both methods agree: $1680$.

Final answers

• (a) Total arrangements $= \boxed{15120}$
• (b) The three Ss together $= \boxed{1260}$
• (c) Starting with A $= \boxed{1680}$