Question

How many numbers can be formed with the digits 1 2 3 4 3 2 1 so that odd digit always occupy the odd places

Answer 1

There are many many numbers can be formed by using these numbers.

Some of them are:

1234321
1234312
1234132
1234123
1234213
1234231
1232431
1232413
1232341
1232314
1233214
1233241
1233421
1233412
1243312
1243321


Thanks

Answer 2

Given Numbers = 1, 2, 3, 4, 3, 2, 1

Number of  Numbers(Total Numbers) = 7

We  can form 7 digits  numbers.

$\framebox{1th}\framebox{2th}\framebox{3th}\framebox{4th}\framebox{5th}\framebox{6th}\framebox{7th}$

Here, Odd  places are  1th , 3th , 5th , 7th

It  means  there are 4 odd  places  in which we have  to fill odd  numbers.

In Given question odd numbers  are = 1, 3, 3, 1

Now,

4 odd places , 4 numbers. Right!!! But here  1 and  3  both are  2   times.

Conclusion we get,

We  have  to arrange 4 numbers  on 4 places  where  1 is  two and  3  is  two times.

So,

Number  of  places  to arrange  them = $=\frac{4!}{2!\times2!}=\frac{4\times3}{2}=6\;ways$

Still even places are  not  filled!

We  have  3  even places = 2th , 4th, 6th

Even numbers are (Given in question)  = 2, 4 , 2

Conclusion is  that , we have  to arrange 3 numbers  on  3  places ,where  2  are  same*(2 , 2 = two times)

Number  of ways  to arrange  them = $\frac{3!}{2!}=\frac{3\times2}{2}=3\;ways$

According  to fundamental principal,

Total number  of  ways = 3 × 6 = 18 ways!