Question

How many 5 digit (numerical) passwords are possible for the school email system if the first digit cannot be 0?​

Answer 1

total numbers available are 10 ...

0,1,2,3,4 5,6,7,8,9...

since 0 CAN'T be first digit ...

THEREFORE there are

9 ×10×10×10×10 = 90000 ways .....if repetition of digits is allowed .....

and if repition of digits is not there are ...

9 × 9 x8x7x6 = CALCULATE............. ...

hope it helps

Answer 2

Given:

Number of digits in the p-assword=5

The first digit cannot be 0

To find:

The number of possible p-asswords

Solution:

The number of possible p-asswords is 1,17,216.

We can find the number by following the given process-

We know that the p-asswords contain digits from 0 to 9.

In case I, the digits of the p-assword cannot be repeated.

The number of digits in the p-assword=5

The number of digits that can be put in the first place=9 (The first digit cannot be 0)

The number of that can be put in the second place=9 (0 can be included)

Similarly, the number of digits for the next place keeps on decreasing by 1.

So, the number of possible p-asswords with no repetition of digits= 9×9×8×7×6

=27,216 p-asswords

In case II, the digits of the p-assword can be repeated.

The number of digits in the p-assword=5

The number of digits that can be put in the first place=9 (The first digit cannot be 0)

The number of that can be put in the second, third, fourth, fifth place=10

So, the number of possible p-asswords with repetition of digits= 9×10×10×10×10

=90,000 p-asswords

The total number of possible p-asswords= P-asswords with digits repeated+ p-asswords with digits not repeated

=90,000+27,216

=1,17,216

Therefore, the number of possible p-asswords is 1,17,216.