A) \[3!4!\]
B) 34
C) 30
D) 12
Correct Answer: C
Solution :
| Out of digits 1, 2,3, 3, 3,2, 1 a seven digit number can be formed with 1st, 3rd, 5th and 7th odd places. |
| Further we have 5 odd digits i.e., 1, 3,3,3,1. |
| Thus arise two cases. |
| Case (i): Using 1,3,3,3 for odd places |
| Number of ways \[=\frac{4!}{3!}\times \frac{3!}{\underbrace{2!}_{\begin{smallmatrix} (\operatorname{for}\, \\ 2,2,1) \end{smallmatrix}}}\] |
| Case (ii): Using 1, 1, 3, 3 for odd places |
| Number of ways\[=\frac{4!}{2!2!}\times \frac{3!}{\underbrace{2!}_{\begin{smallmatrix} (\operatorname{for}\, \\ 2,2,3) \end{smallmatrix}}}\] |
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