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Q.
Number of numbers can be formed by using all the digits $1,2,3,4,3,2, 1$, so that odd digit always occupy odd places, is
Permutations and Combinations
Solution:
Given digits $1, 2, 3, 4, 3, 2, 1$
We need, $7$ digit number, which is same as filling $7$ places in a row with the $7$ digits $1, 2, 3, 4, 3, 2, 1$. Odd places are to be filled by $1, 3, 5, 7$ digits. Odd digits are available
$1, 3, 3, 1$ i.e. two $1 ’s$ and two $3’s$.
$\therefore 4$ odd digits can be arranged at $4$ odd places in $\frac{\,{}^4P_4}{2!2!}$
$ = 6$ ways.
Now, there are $3$ even places.
$\therefore 3$ even digits can be arranged in $ \frac{3!}{2!}$ ways. (As there is two $2's$)
$\therefore $ Total number of ways are $6 \times 3 = 18$