Mathematics - Permutation & Combination Question with Solution | TestHub

In 8 digits numbers, 4 places are odd places. Also, in the given 8 digits, there are three odd digits 1,1 and 3. No. of ways three odd digits arranged at $$\text{ four even places }=\frac{4P_3}{2!}=\frac{4!}{2!}$$ No. of ways the remaining five digits 2 , 2, 2, 4 and 4 arranged at remaining five places $=\frac{5!}{3!2!}$ Hence, required number of 8 digits number $=\frac{4!}{2!}\times \frac{5!}{3!2!}=120$