Answer

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**Hint**: From the question, it is given that all the digits are used \[1,1,2,2,2,2,3,4,4\] and all the digits are taken at a time. So, we should form 9-digit numbers. In these digits, we should find the number of odd digits and number of even digits. We were also given that the odd digits occupy even places. In a 9-digit number, we should find the number of odd digits and number of even digits. We know that the number of ways to place r objects in n places is equal to \[^{n}{{C}_{r}}\left( n\ge r \right)\]. We know that the number of ways to arrange n objects among those n objects r objects are similar and p objects are also similar and (n-r-p) objects are different is equal to \[\dfrac{n!}{r!p!}\]. By using these concepts, we should find the required number of ways.

**:**

__Complete step-by-step answer__From the question, it is given that all the digits are used \[1,1,2,2,2,2,3,4,4\] and all the digits are taken at a time. So, we should form 9-digit numbers. In these digits, we were having 3 odd digits and 6 even digits. We were also given that the odd digits occupy even places. In a 9-digit number, we will have 4 odd places and 5 even places. So, we should place 3 odd digits in 4 even places. We know that the number of ways to arrange n objects among those n objects r objects are similar and (n-r) objects are different is equal to \[\dfrac{n!}{r!}\].

We know that the number of ways to place r objects in n places is equal to \[^{n}{{C}_{r}}\left( n\ge r \right)\].

So, now we should place 3 odd digits in 4 even places.

The number of ways to place 3 odd digits in 4 even places is equal to \[^{4}{{C}_{3}}\].

Now we should find out the number of ways to arrange these 3 odd digits.

We know that the number of ways to arrange n objects among those n objects r objects are similar and (n-r) objects are different is equal to \[\dfrac{n!}{r!}\].

Now we should place 3 odd digits in which two odd digits are similar.

So, the number of ways to place 3 odd digits in which two odd digits are similar is equal to \[\dfrac{3!}{2!}\].

Now we should place remaining numbers.

Now we are having 6 even digits. Now these digits should be arranged.

We know that the number of ways to arrange n objects among those n objects r objects are similar and p objects are also similar and (n-r-p) objects are different is equal to \[\dfrac{n!}{r!p!}\].

Now we should place the remaining 6 digits in which 4 digits and the remaining 2 digits are also similar.

So, the number of ways to place 6 digits in which 4 digits are similar and other 2 digits are similar is equal to \[\dfrac{6!}{4!2!}\].

So, let us assume the number of ways are equal to T.

\[\Rightarrow T{{=}^{4}}{{C}_{3}}\dfrac{3!}{2!}\dfrac{6!}{4!2!}\]

We know that \[^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\].

\[\begin{align}

& \Rightarrow T=\dfrac{4!}{3!\left( 4-3 \right)!}\dfrac{3!}{2!}\dfrac{6!}{4!2!} \\

& \Rightarrow T=\dfrac{4!}{3!1!}\dfrac{3!}{2!}\dfrac{6!}{4!2!} \\

& \Rightarrow T=\dfrac{4.3!}{3!}\dfrac{3.2!}{2!}\dfrac{6.5.4!}{4!2!} \\

& \Rightarrow T=\left( 4 \right)\left( 3 \right)\left( 15 \right) \\

& \Rightarrow T=180 \\

\end{align}\]

So, we can say that the number of such numbers in which the odd digits occupy even places is equal to 180.

**Note**: Some students have a misconception that the number of ways to arrange n objects among those n objects r objects are similar and (n-r) objects are different is equal to \[\dfrac{n!}{(n-r)!}\]. The number of ways to arrange n objects among those n objects r objects are similar and (n-r) objects are different is equal to \[\dfrac{n!}{r!}\]. If this misconception is followed, then the solution may go wrong.

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