Out of the six digits 1, 2, 3, 4, 5 and 6; how many two digit numbers can be formed without repetition of digits ?
- (a)6
- (b)15
- (c)30
- (d)40
Correct — C, 30. A two-digit number needs a tens digit and a units digit. Any of the six digits can occupy the tens place, and since repetition is barred, five remain for the units place: 6 × 5 = 30. None of the six digits is 0, so no arrangement is disqualified for having a leading zero. In permutation notation this is ⁶P₂ = 6! ÷ 4! = 30.
- (a)6 — 6 is the number of digits available, not the number of numbers that can be built from them.
- (b)15 — 15 is ⁶C₂, the number of unordered pairs. Here 12 and 21 are different numbers, so order counts and the answer is twice as large.
- (d)40 — 40 exceeds the total of 36 that even unrestricted repetition would allow, since 6 × 6 = 36.
Counting arrangements starts with a decision about whether order matters. When it does, the count is a permutation: filling r places from n distinct objects without repetition gives n × (n − 1) × … × (n − r + 1), written ⁿPᵣ. When order does not matter the count is a combination, ⁿCᵣ, smaller by a factor of r!.
The single test that decides this item is whether 12 and 21 count as two numbers. They do, so the answer is a permutation and not a combination — which is why 15 sits in the option list. The second habit worth building is checking for zero. Had the digit set included 0, the tens place would have had only five candidates and the count would have dropped to 5 × 5 = 25.
- ⁶P₂ = 6 × 5 = 30 two-digit numbers without repetition.
- ⁶C₂ = 15, which counts unordered pairs and halves the permutation count.
- With repetition allowed the count would be 6 × 6 = 36.
- A digit set containing 0 reduces the choices for the leading place by one.
- For r places from n objects without repetition, the count is n!/(n − r)!.
No zero appears in the set, so every one of the thirty arrangements is a genuine two-digit number.
- Using combinations and answering 15 when order plainly matters.
- Allowing repetition and answering 36.
- Applying the leading-zero correction when no zero is in the given set.
A basic counting item, set as the entry point to the permutation-versus-combination distinction that later questions in the same block build on.
There is a group of 5 people among which there is one couple. In how many ways can these 5 people be seated in a row having 5 chairs if the couple is to be seated next to each other?
- (a) 24
- (b) 48
- (c) 60
- (d) 120
Answer(b) 48
Counting arrangements on an earlier CAPF paper, with a restriction attached. Both items start from the same question — how many ways can these places be filled in order — before any condition is applied.
- practice — not a real PYQ
How many three-digit numbers can be formed from 1, 2, 3, 4 and 5 without repeating a digit?
- (a)10
- (b)20
- (c)60
- (d)125
Answer(c) 60 — 5 × 4 × 3 = 60, that is ⁵P₃.
- practice — not a real PYQ
How many two-digit numbers can be formed from 0, 1, 2 and 3 without repetition?
- (a)6
- (b)9
- (c)12
- (d)16
Answer(b) 9 — the tens place cannot be 0, leaving 3 choices there and 3 for the units place.