Permutation involving different outcomes

Question

So say I have want to find, with the lowercase alphabet, how many strings of 9 different letters that can be formed that have the letter z in front, a block of letters xy (among other random distinct letters), a vowel (a, e, i, o, u) at the end.
This is my working:
There is only one way to pick 1 letter, so that would be for the case for z.
With 7 letters remaining for the string of length 9, and since three predetermined letters are among the 9, then it's just 23 letters than remain and with the permutation:
$P(23, 7) = frac{23!}{(23-7)!}$
The last vowel is simply $P(5,1) = frac{5!}{(5-1)!}$
And the result is the product of those 3. Is that correct?

user2661923 · Accepted Answer

I am not sure what three factors the query is referring to, since I only saw two factors: P(23,7) and P(5,1).
Anyway, this is how I would calculate it.
You have $P_1 = 5$ possibilities for the rightmost position.
Assume that "a" is chosen for that.
Then, the letters x,y,a,z are taken.
There are $P_2 = binom{22}{5}$ ways of selecting the 5 other letters needed.
Here, I am going to assume that the (y,x) block is not allowed, and that instead, the letters must appear in the order (x,y).  If this assumption is false, multiply my final answer by 2.
Finally, since the (x,y) letters must appear as a block, you can construe them as a unit.  Then you have 6 units to be permuted.
This can be done in $P_3 = 6!$ ways.
Final answer is :
$$P_1 times P_2 times P_3 = 5 times binom{22}{5} times 6!.$$

Permutation involving different outcomes

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