Question

In an exactly 8 character long password where capital letters, small letters, and digits (0 to 9) must be used. Regardless of the order, how many passwords will use exactly 5 ones, 4 twos, and one Z?

Answer 1

solution:

Given that

No.of capital letters = 26

No.of small letters = 26

No.of digits(0-9) = 10

Total No.of characters = 26+26+10 = 62

i) No.of passwords will use exactly 5 one's

---> Regardless of the order,

ex-1 :

1

1

1

1

1

ex-2:

1

1

1

1

1

The no.of possible ways to keep 5 one's in password = 8C5 = 56

The no.of possible ways possible ways to keep remaining 3 characters = 61*61*61 [ since 1 is not concerned here and we may keep any character from remaining 61 characters at each digit] By product rule:

   Total No.of passwords that contains exactly 5 ones = 56 * 61 * 61 * 61 = 56*61^3

ii) No.of passwords will exactly use 4 two's  

Regardless order,  

2

2

2

2

The no.of possible ways to keep 4 two's in password = 8C4 = 70

The no.of possible ways possible ways to keep remaining 4 characters = 61*61*61*61 [ since 2 is not concerned here and we may keep any character from remaining 61 characters at each digit]

By product rule:

   Total No.of passwords that contains exactly 4 two's = 70 * 61 * 61 * 61 * 61 = 70*61^4

iii) No.of passwords will use exactly one Z(capital)

   The no.of possible ways to keep one Z in password = 8C1 = 8

The no.of possible ways to keep remaining 7 characters = 61^7   [ since Z(capital) is not concerned here and we may keep any character from remaining 61 characters at each digit]

By product rule:

   Total No.of passwords that contains exactly one Z = 8 * 61 * 61 * 61 * 61 * 61 * 61*61 = 8 * 61^7

The Total No.of passwords will use exactly 5 one's, 4 two's and one Z = (56 * 61^3) + (70*61^4) + (8*61^7)