Question

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:

Answer 1

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 $${}^n{C}_r(n\ge r)$$. 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 $${\displaystyle \frac{n!}{r!p!}}$$. 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 $${\displaystyle \frac{n!}{r!}}$$.

We know that the number of ways to place r objects in n places is equal to $${}^n{C}_r(n\ge r)$$.
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 $${\displaystyle \frac{n!}{r!}}$$.
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 $${\displaystyle \frac{3!}{2!}}$$.
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 $${\displaystyle \frac{n!}{r!p!}}$$.
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 $${\displaystyle \frac{6!}{4!2!}}$$.
So, let us assume the number of ways are equal to T.
$$\Rightarrow T{=}^4{C}_3{\displaystyle \frac{3!}{2!}}{\displaystyle \frac{6!}{4!2!}}$$
We know that $${}^n{C}_r={\displaystyle \frac{n!}{r!(n-r)!}}$$.
$$\begin{array}{rl}& \Rightarrow T={\displaystyle \frac{4!}{3!(4-3)!}}{\displaystyle \frac{3!}{2!}}{\displaystyle \frac{6!}{4!2!}}\\ & \Rightarrow T={\displaystyle \frac{4!}{3!1!}}{\displaystyle \frac{3!}{2!}}{\displaystyle \frac{6!}{4!2!}}\\ & \Rightarrow T={\displaystyle \frac{4.3!}{3!}}{\displaystyle \frac{3.2!}{2!}}{\displaystyle \frac{6.5.4!}{4!2!}}\\ & \Rightarrow T=\left(4\right)\left(3\right)\left(15\right)\\ & \Rightarrow T=180\end{array}$$
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 $${\displaystyle \frac{n!}{(n-r)!}}$$. 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 $${\displaystyle \frac{n!}{r!}}$$. If this misconception is followed, then the solution may go wrong.