Formation of numbers

Analysis 1 $\text{In forming a number, we must}$

$\text{(a) select a group of three odd digits from (1,3,5,7 and 9),}$

$\text{(b) select a group of two even digits from (2,4,6 and 8), and then}$

$\text{(c) form a permutation of the five digits selected in (a) and (b).}$

$\text{The number of ways of performing (a) is 5C3 or 10 ways and for (b) is 4C2 or 6 ways.}$ $\text{For each way of performing (a) and (b) , the number of ways of performing (c) is 5! ways.}$ $\text{Hence, by principle for successive events, the number of ways for performing (a), (b) and (c) is (10)(6)(5!)}=$ $\boxed{\large\color{#D61F06}7200}$

Analysis 2

$\text{There are 5 odd digits and 4 even digits form 1 to 9. We can choose any of the 5 odd digits for the first digit of the number.}$ $\text{Then any of the 4 remaining odd digits for the second digit of the number and any of the 3 remaining odd digits for the third}$ $\text{digit of the number. We can choose any of the 4 even digits for the fourth digit of the number. Then any of the 3 remaining}$ $\text{even digits for the last digit of the number. Now, we must arrange the digits. This can be done in 5! ways, but since there are}$ $\text{3 odd and 2 even, we must divide 5! by 3!2!. So we have}$

$5(4)(3)(4)(3) \frac{5!}{3!2!} =$ $\boxed{\large\color{#D61F06}7200~numbers}$