A password of length 7 is to be generated which contains only two of the letters from A-Z.. One letter repeated $3$ times and the other letter repeated $4$ times. How many such passwords are there?
Can someone please explain to me why $26^7\cdot 7!$ is wrong? Wouldn't $26^7$ be the amount of passwords available from any letter of the alphabet and $7!$ be the number of ways of arranging those letters in the password? So the total number of passwords that can be generated is the product of these two?
 A: The number of ways to choose $2$ out of $26$ letters is:
$$\binom{26}{2}=325$$
The number of ways to choose $3$ places for the $1$st letter and $4$ places for the $2$nd letter is:
$$\binom{7}{3}\cdot\binom{4}{4}=35$$
The number of ways to choose $4$ places for the $1$st letter and $3$ places for the $2$nd letter is:
$$\binom{7}{4}\cdot\binom{3}{3}=35$$
So the total number of passwords is:
$$325\cdot(35+35)=22750$$
A: There are $\binom{7}{3}$ ways to choose 3 from 7, which is 35. The first letter can be one of 26, the second one of 25 (unless both letters are allowed to be the same!). So the answer is $35*26*25=22750$.
A: Well , the answer is 
$$(26*25)*(\frac{7!}{3!*4!})$$
Now , why would your answer be wrong  ? cause you are not considering that you can you can choose only 2 at a time , then the rearrangement depends on only what number you choosed for the first time . after you choose 2 letters , it is the problem how 4 and 3 similar things can be arranged. 
if you consider $26^7$ , here is what happening,
The first place can be filled by 26 letters , second place also 26 , third place also , so you get a combination like ABCDEFG which is not correct. $26^7$ , is all the permutation you can get from 26 letters , when you multiply it by 7! I don't know what you get.
A: Another way to think is -
No of ways of selecting one alphabet out of 26 is $\binom {26}{1}$.
Then the number of ways of arranging it in 3 places out of 7 is $\binom {7}{3}$.
Then we are left with 25 alphabets and we have to select one. This can be done in $\binom{25}{1}$ ways. And we have only one way of arranging it. And therefore our final answer is $$\binom{26}{1}\binom{7}{3}\binom{25}{1}$$.
A: $26^7\cdot 7!$ is clearly way too big. The $26^7$ factor all by itself counts all 7-letter passwords made up from a 26-letter alphabet. Multiplying by $7!$ can only make the number bigger (a lot bigger), when the correct number should be a lot smaller because of all the incorrect combinations.
There are 26 ways to choose the letter that occurs 4 times, and having chosen it there are 25 ways to choose the letter that occurs 3 times. There are $7 \choose 4$  ways to pick the places where the first letter occurs. The total number of passwords is then $26 \cdot 25 \cdot {7 \choose 4}$
