Question

The different six digit numbers whose $3$ digits are even and $3$ digits are odd is

• A. $281250$
• B. $281200$
• C. $156250$
• D. None of these

Answer 1

Answer: (A)

If first digit is even then select other places from remaining five places for two more even digits. This can be done by $4\times {}^5{C}_2\times 5\times 5$ 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 $=4\times {}^5{C}_2\times 5\times 5\times 5^3=8\times 5^6$
$(0$ 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
$=5\times {}^5{C}_2\times 5\times 5\times 53=5\times 10\times 55=10\cdot 5^6$
$\therefore$ Total six digit numbers in which $3$ odd and $3$ even are
$=8\times 5^6+10\times 5^6=18\times 5^6=281250$