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Q.
The different six digit numbers whose $3$ digits are even and $3$ digits are odd is
Permutations and Combinations
Solution:
If first digit is even then select other places from remaining five places for two more even digits. This can be done by $4 \times \,{}^5C_2 \times 5 \times 5$ ways. Now, the remaining three places are to be fit up by odd digits, which can be done in $5^3$ ways. In this way, number of six digits numbers in which three digits are even and $3$ are odd and first digit
is even $= 4 \times \,{}^5C_2 \times 5 \times 5 \times 5^3 = 8 \times 5^6$
$(0$ can’t occur at first place)
Similarly, number of six digit numbers, which begin with an odd number/digit and which have $3$ even, $3$ odd digits
$= 5 \times \,{}^5C_2 \times 5 \times 5 \times 53 = 5 \times 10 \times 55 = 10 \cdot 5^6$
$\therefore $ Total six digit numbers in which $3$ odd and $3$ even are
$= 8 \times 5^6 + 10 \times 5^6 = 18 \times 5^6 = 281250$