Mathematics - Permutation & Combination Question with Solution | TestHub

Odd digits:$1, 1, 3$
Even digits:$2, 2, 2, 2, 4, 4$
The number of ways of placing odd digits at even position$= {}^{4}C_3\times \frac{3!}{2!}=4\times 3=12$
The number of ways of placing even digits$=\frac{6!}{4! 2!}=15$
$\therefore$Total number of ways$=12\times 15=180$