# User Forum

Subject :IMO    Class : Class 3

How many four digit numbers can be formed, using 7, 5, 0, 2 only once in a number?

A 4
B 12
C 9
D 18

Do we have formulae to solve this problem?

Class : Class 3

Class : Class 3

## Ans 3:

Class : Class 1
its so easy right

## Ans 4:

Class : Class 3
lol class 1 ha ha ha..

## Ans 5:

Class : Class 3
n!-(n-1)! is the answer. n is number of digits here. (n-1)! is subtracted from n! since there is digit with 0. otherwise it is just n! Now How do you calculate n!. For example 4! here. 4! = 4x3x2x1 = 24 (4-1)! = 3! = 3x2x1 = 6 Hence the answer is n! - (n-1)! = 4! - (4-1)! = 4! - 3! = 24 -6 = 18 Option D is the answer.

Class : Class 3
(D) 18

## Ans 7:

Class : Class 6
Using Factorial for Class 3 seems difficult. I taught my daughter as under:- Step 1 - Put 7 at thousand place and how many different numbers we can get --She replied 6 Step 2- Put 5 at thousand place and how many different numbers we can get --She replied 6 Step 3 -Put 0 at thousand place , do we get four digit number --She replied NO Ste 4 - Put 2 at thousand place and how many different numbers we can get --She replied 6 Step 5 - Add numbers obtained in above steps we get 18

## Ans 8:

Class : Class 4
TOUGH FOR CLASS 1