CSAT Solved Papers/ 2023/Q19

2023 CSAT — Q19

Quant Counting & combinatorics 2.5 marks Medium

How many distinct $8$ -digit numbers can be formed by rearranging the digits of the number $11223344$ such that odd digits occupy odd positions and even digits occupy even positions?

A 12
B 18
C 36 Answer
D 72

Worked rationale

Split the digits by parity and the positions by parity, then count each block independently and multiply.

The digits of $11223344$ are: odd digits $\{1,1,3,3\}$ and even digits $\{2,2,4,4\}$ .

There are $4$ odd positions ( $1,3,5,7$ ) and $4$ even positions ( $2,4,6,8$ ).

Place the odd digits $\{1,1,3,3\}$ in the $4$ odd positions: arrangements $= \dfrac{4!}{2!\,2!} = 6$ .
Place the even digits $\{2,2,4,4\}$ in the $4$ even positions: arrangements $= \dfrac{4!}{2!\,2!} = 6$ .

The two placements are independent, so

6 \times 6 = 36.

Answer: (c) 36.

Why the other options miss

A

added instead of multiplied: computes $6 + 6$ (adds the two blocks) instead of multiplying the independent placements.
B

one block divided wrongly: mis-divides one block’s count (e.g. $4!/2! = 12$ for evens but $6$ for odds, $12 \cdot 6$ mis-simplified), or handles only one repeated pair.
D

forgot a repeated-digit division: leaves out the repeat division on one side ( $4! = 24$ for one block, $24 \times 3$ ), over-counting swaps of identical digits.

Specialist insight

The parity constraint decouples the problem into two independent multiset permutations. Each is $\dfrac{4!}{2!\,2!} = 6$ because of the two repeated digits, and the multiplication principle gives $6 \times 6 = 36$ . The two traps are (i) forgetting the $2!2!$ repetition division (gives $24$ or $72$ ), and (ii) adding instead of multiplying. Both blocks must be handled with repetition and then multiplied.

The trap, in one line

Odd digits in odd slots $=\frac{4!}{2!2!}=6$ , even in even $=6$ , independent $\Rightarrow 6\times 6 = 36$ (multiply, divide by the repeats).

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