Question 406031

A) (i) for first and last position any digit except 0, for other any of 10 digits 

 total no. of possible numbers = 9*10*10*10*10*10*10*9 = 81000000 = 81*10^6

(ii)
for odd numbers odd number must in at unit place....

total no. of odd numbers = 5
we can put digit 2 at first place by 1 way.

total no. of possible numbers = 1*10*10*10*10*10*10*5 = 5000000 = 5 * 10^6




B) if repetitions are not allowed...


(i) as first place and last can't be zero, 

 no. of ways to put digit at first and last position = 9*8 = 72 
 
( now we have 8 digits and 6 positions to fill ) 

no. of ways to put digit at remaining 6 positions = 8P6 = 20160 

 total no. of ways = 72*20160 = 1451520