We have 7 numbers: 1-2-3-4-5-6-7, out of which 4 are odd. Want to make 5 digit code, plus first and last digits must be odd: OXXXO.

#1:
\(P^2_4*P^3_5=12*60=720\)

\(P^2_4\) - choosing 2 odds out of 4 odds, (note that order matters in this case: 1---3 is different from 3---1);

\(P^3_5\) - choosing 3 numbers (for XXX) out of 5 left, (order matters: o234o is different from o342o).

OR #2:
4 choices for the first slot, 3 choices for the last slot (4 odds-1=3), 5 choices for second slot (7 numbers -2 odds), 4 choices for the third slot and 3 choices for the fourth slot: 4*3*5*4*3=720.

OR #3:
\(C^2_4*2!*C^3_5*3!=720\), this is basically the same as the first approach, but here we choose 2 odds out of 4 and then arranging them (2!) and respectively we choose 3 other numbers from 5 left and then arranging them (3!).

Hope it's clear.
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720 is the answer.

They use a "slot technique":

Ie. Spaces _,_,_,_,_

Space one has 4 options (1,3,5,7) and the last space has 3 (the 3 non-chosen odds from space one)

So we're sitting at 4,_,_,_,3

So between the first and last number been chosen 2 out of the 7 possibilites have been used leaving us with

4*5*4*3*3=720 combos

Ahhhhh, my problem on the related question is that I assumed the numbers were 1-9 when it was actually 0-9.

FRIG!!!

Thanks for the help guys!!
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