goyalsau wrote:A has to create a password having 5 distinct characters using at least 2 digits (from 7 to 9) and at least two english vowels (from A, E, I, O, U). No character other than digit from 7 to 9 and vowels from english alphabets are allowed in that password. When the password starts with a digit it must end with an alphabet and when it starts with an alphabet it must end with a digit. Find the number of possible passwords that A can create.
(a) 1260
(b) 4480
(c) 3620
(d) 2880
_ _ _ _ _
1. We use 2 digits and 3 alphabets:
a. We use first as digit and last as alphabet
first can be chosen in = 3 ways (digit)
last can be chosen in = 5 ways (alphabet)
In remaining 3 places 1 digit out of remaining 2 digits can be chosen in 3*C(2,1) = 6 ways
In remaining 2 places 2 alphabets out of remaining 4 alphabets can be chosen in 2*C(4,2) = 12 ways
Total = 3*5*6*12 = 1080
b. We use first as alphabet and last as digit
last can be chosen in = 3 ways (digit)
first can be chosen in = 5 ways (alphabet)
In remaining 3 places 1 digit out of remaining 2 digits can be chosen in 3*C(2,1) = 6 ways
In remaining 2 places 2 alphabets out of remaining 4 alphabets can be chosen in 2*C(4,2) = 12 ways
Total = 3*5*6*12 = 1080
2. We use 3 digits and 2 alphabets:
a. We use first as digit and last as alphabet
first can be chosen in = 3 ways (digit)
last can be chosen in = 5 ways (alphabet)
In remaining 3 places 1 alphabets out of remaining 4 alphabets can be chosen in 3*C(4,1) = 12 ways
In remaining 2 places 2 digits out of remaining 2 digits can be chosen in 2*C(2,2) = 2 ways
Total = 3*5*12*2 = 360
b. We use first as alphabet and last as digit
last can be chosen in = 3 ways (digit)
first can be chosen in = 5 ways (alphabet)
In remaining 3 places 1 alphabets out of remaining 4 alphabets can be chosen in 3*C(4,1) = 12 ways
In remaining 2 places 2 digits out of remaining 2 digits can be chosen in 2*C(2,2) = 2 ways
Total = 3*5*12*2 = 360
Total number of possible passwords = 1080+1080+360+360 = 2880
