How many integer can be formed using 2,3,5,9 and 3 as digit;? A/if repetition is not allowed B/iF no consecutive digits are the same
Answers
Answer:
Step-by-step explanation:
solution:
1. 3 digits formed from 2, 3, 5, 6, 7, 9
Permutation formula =
(n−r)!
n!
=
(6−3)!
6!
3
6
P=66×5×4
=120
2. Numbers less than 400 can start with digits 2 and 3 only.
2×6×6=72
3. Even numbers are 2 and 6 from the given 6 digits.
2×6×6=72
4. Odd numbers are 3, 5, and 7 from the given 6 digits.
6×6×3=108
5. 5 is the only digit from 6 digits have multiples of 5.
6×6×1=36.