[2375] How many five digit numbers are there such that two left most digits are even and remaining are odd

For Competitive Exams

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

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