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# All possible numbers are formed using the digits $1,1,2,2,2,2,3,4,4$ taken all at a time. The number of such numbers in which the odd digits occupy even places is:

Last updated date: 13th Jun 2024
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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.

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.