Using the numbers 0,1,2,...,9 as digits, how many 4-digit numbers

Gybrisysmemiau7

Gybrisysmemiau7

Answered question

2022-06-21

Using the numbers 0,1,2,...,9 as digits, how many 4-digit numbers exist in which all digit are different with two digits as even numbers and two as odd numbers?

Answer & Explanation

Daniel Valdez

Daniel Valdez

Beginner2022-06-22Added 19 answers

First pick 2 odd numbers and 2 even numbers. There are ( 5 2 ) 2 ways to do this. Now order them. There are 4! ways. So there are 4 ! ( 5 2 ) 2 total numbers, but we have included 4-digit numbers that start with 0. Now let's count the number that start with 0. There are 4 choices for the other even digit, ( 5 2 ) choices for the odd digits, and 3! orders. So for our final answer we have:
4 ! ( 5 2 ) 2 3 ! ( 4 ) ( 5 2 )
Arraryeldergox2

Arraryeldergox2

Beginner2022-06-23Added 10 answers

Another possibility would be this : consider the 4-digit numbers without 0 in them. There is 4 possible even numbers and 5 possible odd numbers, so ( 5 2 ) ( 4 2 ) = 60 possibilities. There are 4 ! = 24 ways to place them, hence you have 60 × 24 = 1440 such numbers.
Now consider those with 0 in them. The zero must be one of the 3 first digits. Now for the number possibilities, there are 4 choices for the even number and ( 5 2 ) = 10 choices for the odd numbers, hence 40 choices for the numbers. Afterwards, there are 3 possibilities for the position of 0 and 3 ! = 6 choices for the position of the three other digits, hence 18 possibilities in total. Therefore there is 40 × 18 = 720 such digits.
Summing up gives 1440 + 720 = 2160 = 4 ! ( 5 2 ) 2 3 ! ( 4 ) ( 5 2 ) possibilities.

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