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Numbers are formed using digits $1,1,2,2,2,2,3,4,4$ taken all at a time. Find number of such numbers so that odd digits occupy even places.

The odd digits are $1,1,3$ and there are 4 even places

So permutation of arranging 3 items in 4 places where two items are alike is.

Now the solution given to me involves use of combinatorics, and I can’t understand how they came up with it, since we are arranging and not selecting. Can I get an explanation on how combinations can be used to solve this (or how to do it without combinations)

Aditya
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2 Answers2

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Total numbers which can be formed : $ \cfrac {9!}{2!4!2!} $

Total odd digits : $3 , i.e. 1,1,3 $

Total even places : $4$

Ways in which you can arrange the digits : $⁴C_3 * \cfrac{3!}{2!}$ $\longrightarrow (1)$

$3! - $total ways of arranging $ 1,1,3$

$2! - $two $1's$ are alike

Of course, there is no restriction on the other digits,

Ways of arranging the other digits : $ \cfrac {6!} {4!2!}$ $\longrightarrow (2)$

$6! - $total ways of arranging $ 2,2,2,2,4,4 $

$2! - $two $4's$ are alike

$4! - $four $2's$ are alike

Now you just multiply $(1)$ and $(2)$ on account of multiplication principle.

The final answer is : $⁴C_3 * \cfrac{3!}{2!}$ $*$$ \cfrac {6!} {4!2!}$

Hope this clears your doubt.

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All possible numbers that can be formed = $9!$/$2!4!2!$

No. of numbers such that odd digits occupy even places = $⁴C_3$×($3!$/$2!$)×($6!$/$4!2!$)

Sangeeta
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