probability in permutations how many different 7-place liscence plates are possible if the third three places are occupied by letters and final 4 by numbers.
the answer I found is 
26.26.26.10.10.10.10 --175, 760, 00
I really did not understant how it come. can someone help me
 A: The first place of the license plate can be filled by any of the 26 alphabets.The same holds for the second and the third place.Therefore total no. of ways to occupy the first 4 places = $26\times 26\times 26$
The 4th,5th and 6th can be filled by any of the 10 digits (0,1,2,3,...,9). Number of ways to do so = $10\times 10\times 10\times$
Therefore total number of ways = $26^{3}\times 10^{4}$
A: We are told that a license plate consists of $3$ letters, followed by $4$ digits. So for example $BBA7005$ is a valid license plate. Note that letters and digits can repeat. 
For the detailed analysis, we concentrate on the "letter" part of the license plate. So we ask: how many $3$-letter "words" are there? 
Let's first solve the much simpler problem, how many $1$-letter "words" are there? If we are using the ordinary "English" alphabet, there are $26$.
How many $2$-letter words are there? Let's imagine listing them alphabetically. There are $26$ words that begin with $A$, and $26$ that begin with $B$, and $26$ that begin with $C$, and so on. So in total there are $26+26+26+\cdots+26$ $2$-letter words, where our sum has $26$ terms. More simply, there are $26\times 26=26^2$ $2$-letter words.
How many $3$-letter words are there? Again, imagine listing them alphabetically. There are the words that begin with $A$. We can form such a word by putting after the $A$ any $2$-letter word. We have seen there are $26^2$ $2$-letter words, so there are $26^2$ $3$-letter words that begin with $A$. Similarly, there are $26^2$ $3$-letter words that begin with $B$, and so on, for a total of $26\times 26^2=26^3$ $3$-letter words.
A similar argument shows that there are $10^4$ $4$-digit sequences. These too are "words," over an alphabet that consists of the $10$ digits.
Since we can make a license plate by taking any $3$ letter word, and following it with any $4$-digit "word", there is a total of $(26^3)(10^4)$ possible license plates. 
