How many 4 digit positive integers have exactly 2 odd digits and 2 even digits?
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To start off this problem, you need to separate this into two cases:
Case 1) Thousands digit is odd.
The possibilities for the odd digits are: 1, 3, 5, 7, and 9.
You need to choose 1 of the 5 integers as the first digit. Then you need to choose one slot of the remaining three to put the other odd digit, which is 3 choose 1. After you found out the chosen slot, there are 5 numbers to choose from. Then you are left with 2 even digit slots, which have these digits to choose from: 0, 2, 4, 6, and 8.
Result: 5 C 1 × 3 C 1 × 5 C 1 × 5 C 1 × 5 C 1 = 5 4 × 3 = 1,875
Case 2) Thousands digit is even. This case is similar to the case above, except for the first digit, the digit can not be 0, so there are these numbers to choose from: 2, 4, 6, and 8.
Result: 4 C 1 × 3 C 1 × 5 C 1 × 5 C 1 × 5 C 1 = 5 3 × 3 × 4 = 1,500
Since the relationship between cases is addition, you add 1,500 + 1,875 and the result is 3375.
There are 3 3 7 5 4 digit positive integers that have exactly 2 odd digits and 2 even digits.
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If the first three digits are all odd, or all even, the number cannot contain 2 even and 2 odd digits. To find such numbers:
For each thousands digit, there are 5 hundreds digits which match its parity, and for each of these, there are 5 tens digits which also match parity. As there are 9 possible thousands digits, 9 × 5 × 5 = 2 2 5 0 .
Excluding these leaves us with the numbers with the first three digits consisting of 2 of one parity and 1 of the other. Of these numbers, exactly half contain 2 digits of each parity. (consider: each set of 1 0 integers with a specific first 3 digits contains 5 possible final digits with the correct parity.)
And so, as there are 9 0 0 0 positive four-digit integers, 2 9 0 0 0 − 2 2 5 0 = 3 3 7 5 .