A) 160
B) 120
C) 60
D) 48
Correct Answer: B
Solution :
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 four even places \[=\frac{4{{P}_{3}}}{2!}=\frac{4!}{2!}\] 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 \[=\frac{4!}{2!}\times \frac{5!}{3!2!}=120\]
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