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

Dear student
$$\begin{gathered} Thegivendigitsare:1,2,3,4,3,2,1. \\ Outofthese1,3,3,1areodddigits. \\ Theodddigitsoccupytheoddplacesietheyoccupythe1st,3rd,5thand7thplace. \\ \therefore 4oddplacescanbefilledwith4odddigits=\frac{4!}{2!2!}=\frac{4\times 3\times 2}{2\times 2}=6ways \\ Thethreeevenplacestobefilledwiththeevendigits2,4,2. \\ Theseplacescanbefilled=\frac{3!}{2!1!}=\frac{3\times 2!}{2!}=3ways \\ Hencetherequirednumberofnumbres=6\times 3=18 \end{gathered}$$
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