How many five digit numbers are there such that two left most digits are even and remaining are odd and digit 4 should not be repeated.
2578
2365
2475
2375
Explanation:
We have
4 cases of first digit {2,4,6,8}
5 cases of second digit {0,2,4,6,8}
But 44 is one case we have to omit. So total ways for leftmost two digits are 4 x 5 - 1 = 19
5 cases of third digit {1,3,5,7,9}
5 cases of fourth digit {1,3,5,7,9}
5 cases of fifth digit {1,3,5,7,9}
So total ways = 19 x 5 x 5 x 5 = 2375
We have
4 cases of first digit {2,4,6,8}
5 cases of second digit {0,2,4,6,8}
But 44 is one case we have to omit. So total ways for leftmost two digits are 4 x 5 - 1 = 19
5 cases of third digit {1,3,5,7,9}
5 cases of fourth digit {1,3,5,7,9}
5 cases of fifth digit {1,3,5,7,9}
So total ways = 19 x 5 x 5 x 5 = 2375