As the title states, the question is: "How many ways can six of the letters of the word ALGORITHM be selected and written in a row if the first letter must be A?"

I don't really get what the problem is asking me. I know I have to solve it with permutations, but I really don't know where to start or when to know I've come up with the right answer.

I know it has to be done through permutations. I'm honestly a little confused on the question they are asking though, if it's saying that it must be ALGORI[][][] where three can be anything in which case it would be (9 + 8 + 7 + 6 + 5 + 4) * 3!

Or is it something like P(9,3) = 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 / 6 * 5 * 4 * 3 * 2 * 1??


1 Answer 1


We really need to choose $5$ non-A letters from the $8$ non-A's available, and put them in a row after the A.

The first letter (after the obligatory A) can be chosen in $8$ ways.

For each of these ways, the second letter can be chosen in $7$ ways.

For each choice of first and second, there are $6$ ways to choose the third, and then $5$ ways to choose the fourth, and $4$ ways to choose the fifth, for a total of $(8)(7)(6)(5)(4)$.

Another way: We can choose the $5$ letters from the $8$ available in $\binom{8}{5}$ ways. For every choice, we can line up the $5$ chosen letters in $5!$ ways, for a total of $\binom{8}{5}5!$.

Another way: We do not find the $\text{P}(n,r)$ notation particularly useful. But since it was mentioned in the post, we want to make a $5$-letter word (to append to A) from the $8$ non-A's. There are $\text{P}(8,5)$ ways to do this.


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