SOLUTION: From the digits (1, 2, 3, 4, 5, 6, 7), how many numbers of four different digits each can be formed, if each number involves two odd and two even digits?

Algebra ->  Probability-and-statistics -> SOLUTION: From the digits (1, 2, 3, 4, 5, 6, 7), how many numbers of four different digits each can be formed, if each number involves two odd and two even digits?      Log On


   

Question 885640: From the digits (1, 2, 3, 4, 5, 6, 7), how many numbers of four different digits each can be formed, if each number involves two odd and two even digits?
Answer by Edwin McCravy(20064) About Me  (Show Source):
You can put this solution on YOUR website!
{1, 2, 3, 4, 5, 6, 7} 

Choose a pair of odd digits. 
That is, choose 2 from {1,3,5,7}. C(4,2) = 6 ways

Choose 2 positions for them to go in the 4-digit number. 
That is, choose 2 from {1st,2nd,3rd,4th}.  C(4,2) = 6 ways

Choose the order to put them in the 4-digit number. That is, put the 
smaller odd digit left of the larger odd digit or the larger odd digit 
left of the smaller odd digit.  That's 2 ways.

Choose a pair of even digits. 
That is, choose 2 from {2,4,6}. C(3,2) = 3 ways

There are only two places left for them to go in the 4 digit number.

Choose the order to put them in the 4-digit number. That is, the 
smaller even digit left of the larger even digit or the larger even digit
left of the smaller even digit.  That's 2 ways.

Answer = 6󬝮󫢪 = 432 ways.

Edwin