How many permutations of any 7 letters contain the string bea? Bea has written a computer program that randomly generates 7 letters of the alphabet without replacement, what is the chance a random list will contain her name, such as, 'cbeadfk'. Answer as a fully simplified fraction. 
My solution is:
There are 26 letters in the alphabet and 7 are chosen each time, therefore the number of possible permutations is,
26P7 = 3315312000.
We are interested in strings containing B, E, A in that order so we take those letters out of the alphabet and string, leaving 4 positions with 23 letters to choose from. The string bea can appear in 5 different permutations for each arrangement of the remaining 4 letters. So the amount of strings containing bea is
(23P4)*5 = 1062600
favourable/total = 1062600/3315312000=1/3120
Is my answer correct?
 A: Another way to calculate it:
Choosing $m$ letters in sequence without replacement from the $26$ letters, you have probability $m/26$ to choose the letter B. But in order for the seven letters to include “BEA” the letter B has to be among the first five letters chosen. The total probability of that is $5/26.$
Given that B is among the first five letters, the probability that the next letter is E is $1/25.$ Given B in the first five letters followed by E, the probability that the next letter is A is $1/24.$
So altogether the probability of “BEA” within the first seven letters is
$$
\frac5{26} \cdot \frac1{25} \cdot \frac1{24}
= \frac1{3120},
$$
which agrees with your answer. I also agree that your method is correct. (It is not an accident that both methods get the same answer!)
A: Probability of $bea????$: $\dfrac{1\cdot1\cdot1\cdot23\cdot22\cdot21\cdot20}{26\cdot25\cdot24\cdot23\cdot22\cdot21\cdot20}$.
Probability of $?bea???$ : $\dfrac{23\cdot1\cdot1\cdot1\cdot22\cdot21\cdot20}{26\cdot25\cdot24\cdot23\cdot22\cdot21\cdot20}$.
Probability of $??bea??$ : $\dfrac{23\cdot22\cdot1\cdot1\cdot1\cdot21\cdot20}{26\cdot25\cdot24\cdot23\cdot22\cdot21\cdot20}$.
Probability of $???bea?$ : $\dfrac{23\cdot22\cdot21\cdot1\cdot1\cdot1\cdot20}{26\cdot25\cdot24\cdot23\cdot22\cdot21\cdot20}$.
Probability of $????bea$ : $\dfrac{23\cdot22\cdot21\cdot20\cdot1\cdot1\cdot1}{26\cdot25\cdot24\cdot23\cdot22\cdot21\cdot20}$.
Hence in total,
$$\frac{5}{26\cdot25\cdot24}=\frac1{3120}.$$
