How many five digit natural numbers, with different digits can be formed from the digits 1 to 9 if each number is to contain three odd digits and two even digits?
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even numbers: 2,4,6,8
odd numbers : 1,3,5,7,9
we have to form a number 5 digit number with 3 odd digits and two even digits
now number of ways in which 3 odd numbers can be taken=5C3
number of ways in which 2 even numbers can be taken=4C2
now number of ways of forming the number=5C3*4C2*5!
because each combination obtained can be written in 5! ways