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How many four digit numbers can be formed with the 10 digits $0, 1, 2, \ldots, 9$ if no number can start with 0 and if repetitions are not allowed?

$\_\; \_\;\_\; \_$

Since, number of digits is $4$ the number cannot start with $0.$

At first place all number can appear except $0.$ So, number of possibilities $= 9.$

$9\; \_\;\_\; \_$

Since, repetition is not allowed, second place can be occupied by $8$ different numbers. But here $0$ may appear so total possibilities is $9$.

$9\times9\;\_\; \_$

Similarly, remaining two places can be filled in

$9\times9\times8\times7= 4536.$

1 comment

good ans
At First We Select 10 numbers from 0,1,2.....9 and their combinations that is= 10c4*4!

= 5040

But it will include all those numbers with starting position 0. eg- 0234,0456,0765..... so we have to deduct it

from the total combitations : 5040- 9c3*3!

= 4536.

1 comment

10*9*8*7 - 1*9*8*7 = 9*8*7*9 = 4536. :)