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The letters ABCDEFGH are to be used to form strings of length four. How many strings contain the letter A if repetitions are not allowed?

The answer that I have is :

$$ \frac{n!}{(n-r)!} - \frac{(n-1)!}{(n-r)!} = \frac{8!}{4!} - \frac{7!}{4!} = 8 \times 7 \times 6 \times 5 - (7 \times 6 \times 5) = 1470 $$ strings.

If you could confirm this for me or kindly guide in me the right direction, please do let me know.

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you have 7!/4! ways of having 3 letter strings (we are excluding A). Then insert A in 4 places for each. So, the answer is $7!/4! \times 4 = 840$. Posted this here because I may have made a mistake. – picakhu Dec 2 '11 at 21:46
I see your logic, but do you also see my logic ? I do not know which is right , maybe someone else will get in on this. – Farshid Palad Dec 2 '11 at 22:02
Your second term is incorrect. It should be $(n-1)!/((n-1)-r)!$, not $(n-1)!/(n-r)!$. – Arturo Magidin Dec 2 '11 at 22:04
If we do it your way (which is fine), note that you should be subtracting "$\dfrac{(n-1)!}{(n-1-r)!}$" which is $7\times 6\times 5\times 4$. – André Nicolas Dec 2 '11 at 22:08
up vote 2 down vote accepted

The answer you have is trying to count as follows: there are $8\times 7\times 6\times 5$ strings of four letters from among ABCDEFGH with no repetitions, and no further restrictions.

If we exclude $A$, then we have $7\times 6\times 5\times 4$ strings of four letters from among BCDEFGH with no repetitions.

So the total number that do include A is equal to the total number, minus those that do not contain $A$.

However, the second term for this count is incorrect: you should have $$\frac{(n-1)!}{((n-1)-r)!} = \frac{7!}{3!},$$ not what you have.

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Thank you , I see my mistake, I did 7 choose 3 instead of 7 choose 4. Rookie mistake. – Farshid Palad Dec 2 '11 at 22:07

Fix A first, then the then you have $7$ choices for the remaining $3$ places, then number of possible arrangements: $$7 \times 6 \times 5$$

Now there are exactly $4$ places where the that A possible fit, making the total number of possible arrangements as: $$7 \times 6 \times 5 \times 4 = 840$$

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I presume I am correct. Here is a detailed proof.

First exclude 'A' and permute the rest (7P3). Which can be done in $\frac{7!}{4!}$ ways.

Then, include 'A' back into those permuted cases. $|X_1|X_2|X_3|$ and as indicated by the vertical lines can be in 4 locations. So, the answer is

$$\frac{7!}{4!} \times 4 = 840$$

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There are ${7\choose 3}=35$ ways to choose 3 letters from BCDEFGH. Add an "A" to this set and arrange these four distinct letters in $4!$ ways. This gives $35\cdot 24=840$ strings in total.

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