CSAT Solved Papers/ 2021/Q66

2021 CSAT — Q66

Quant Number theory 2.5 marks Medium

Using $2, 2, 3, 3, 3$ as digits, how many distinct numbers greater than $30000$ can be formed?

A 3
B 6 Answer
C 9
D 12

Worked rationale

A number exceeds $30000$ exactly when its leading (ten-thousands) digit is $\ge 3$ — i.e. the first digit is $3$ (the digits available are only $2$ and $3$ ). Fix a $3$ in front; the remaining four digits are $\{2, 2, 3, 3\}$ , arranged in

\frac{4!}{2!\,2!} = \frac{24}{4} = 6

distinct ways. Each such number starts with $3$ , so it is at least $32233 > 30000$ — all $6$ qualify.

(Numbers starting with $2$ are at most $23333 < 30000$ , so none of those count.)

Answer: (b) 6.

Why the other options miss

A

under-counted the arrangements: of $\{2,2,3,3\}$ , e.g. forgets the repeated-digit division gives $6$ , not $3$ .
C

miscounted a repeated factor: mishandles the repeated digits (treats one pair as distinct), inflating the count.
D

miscounted a repeated factor: divides by only one factorial ( $4!/2! = 12$ ), forgetting the second pair of repeats.

Specialist insight

The threshold " $> 30000$ " is purely a leading-digit condition: only a $3$ in front works, after which the count is the multiset permutation of the rest, $\dfrac{4!}{2!\,2!} = 6$ . The trap is the repeated digits — both the two $2$ s and the remaining two $3$ s must be divided out, or the count balloons to $12$ or $9$ . Fix the constrained position first, then permute the remainder with the correct repetition divisor.

The trap, in one line

Lead with $3$ (else $<30000$ ); arrange $\{2,2,3,3\}$ in $4!/(2!2!) = 6$ ways $\Rightarrow$ (b).

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