A number of four different digits is formed with the help of the digits 1,2,3,4,5,6,7 in all possible ways.
(i) How many such numbers can be formed?
(ii) How many of these are even?

Explanation

Here total digit-7 and no two of which are alike
(i) Required number of ways=Taking 4 out of 7
`=.^(7)P_(4)=7xx6xx5xx4=840`
(ii) For even number of last digit must be 2 or 4 or 6. Now the remaining three first places on the left of 4-digit numbers are to be filled from the remaining 6-digits and this can be done in
`.^(6)P_(3)=6*5*4=120` ways
and last digit can be filled in 3 ways.
`therefore`by the principal of multiplication, the required number of ways
`=120xx3=360`