SOLUTION: please help me how many five digits numbers can be formed from 1,9,4,2 and 7 if 4 and 2 are always to be together.

Your question:
How many five digits numbers can be formed from 1,9,4,2 and 7 
if 4 and 2 are always to be together.

Solution:

I am assuming that each digit can be used only once.

If 4 and 2 are to be always together, consider them as one unit. 
The other 3 numbers form 3 units.

Remember the principles of permutations. 

n things can be arranged in n positions in n! factorial ways. 
The first position can be filled in n ways, the second in (n-1) ways 
and so on. So the total number of ways is n*(n-1)*(n-2)...*1 = n!)

In general, if A can be done in n ways, and B can be done in
m ways, together A *and* B can be done in n*m ways)

Applying the above rules, the 4 units can be arranged in 4! 
(factorial 4) ways.

Now the 2 digits which are in the single unit (2,4) can be themselves 
arranged in 2! or 2 ways. 

So the total number of arrangements = 2! * 4! = 2 * 24 = 48.

The total number of 5 digit numbers (keeping 2 and 4 together) = .

:)