Forming of Numbers

Probability Level pending

From the digits 2,3,4,5,6,7 and 9, how many numbers of four different digits each can be formed, if each number involves two odd and two even digits?


The answer is 432.

A Former Brilliant Member by A Former Brilliant Member

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2 solutions

Nick Schreiner
Jan 4, 2017

There are 4C2 ways to arrange two odd and two even digits. For each of those ways, there are 3P2 ways to place the evens and 4P2 ways to place the odds.

This gives a total of (4C2)(3P2)(4P2)=6*6*12=432.

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This is my solution. I will select a group of 2 odd digits from the 4 odd digits given. Then I will select a group of 2 even digits from the 3 even digits given. Then I will form a permutation of the 4 digits selected. This will give a total of (4C2)(3C2)(4!) = 432

A Former Brilliant Member - 4 years, 5 months ago

From 2 , 3 , 4 , 5 , 6 , 7 , 9 2,3,4,5,6,7,9 , there are 4 4 odd and 3 3 even. The number of ways of choosing 2 2 odd numbers from 4 4 odd numbers is 4 C 2 = 6 4C2=6 and the number of ways of choosing 2 2 even numbers form 4 4 even numbers is 3 C 2 = 3 3C2=3 . The number of ways of arranging these 4 4 numbers is 4 ! = 24 4!=24 . Therefore, we have 6 ( 3 ) ( 24 ) = 432 6(3)(24)=432 different 4 4 digit numbers.

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