8-digit numbers are formed using the digits 1,1,2,2,2,3,4,4.

Question

8-digit numbers are formed using the digits 1,1,2,2,2,3,4,4. The number of such numbers in which the odd digits do not occupy odd places, is:

Accepted Answer

There are 4 odd positions and 4 even positions. Odd digits present: 1,1,3 (three odd digits). Since odd digits must NOT occupy odd places, they must go into the 4 even places.

Arrange 3 odd digits (1,1,3) into 4 even places: $\dfrac{^4P_3}{2!} = \dfrac{24}{2}=12$

Remaining 5 digits (2,2,2,4,4) fill the 5 remaining places (1 leftover even + 4 odd places): $\dfrac{5!}{3!2!}=10$

$$\text{Total} = 12\times10 = 120$$

Solutions · 1