Permutations and Combinations JEE 2026: Why 48, Not 72?

Permutations and Combinations JEE 2026: Why is the answer 48, not 72?

The answer is 48, option B, because input 1 has two available outputs in one case and three in each of the others. This permutations and combinations JEE 2026 question is solved below using the supplied official casewise method.

Take the set:

A = {1, 2, 3, 4, 5, 6} .

Assign each input a distinct output from this same set. The output of 1 must exceed 3, the output of 3 must be below 4, and the outputs of 2 and 3 must add to 5.

This is a single-correct JEE Main 2026 question. “Hard” is the question bank’s classification, not an official NTA difficulty rating.

Which outputs should you fix first?

Fix the two outputs linked by the sum before choosing the output of 1. This shows which allowed outputs have already been used. Six distinct images chosen from a six-element codomain exhaust that codomain, so every one-one function here is a bijection, hence a permutation of the set.

Translate the inequalities and rearrange the sum:

f (1) \in {4, 5, 6}, f (3) \in {1, 2, 3},

$f (2) = 5 - f (3) .$ Evaluate the three possible values separately:

f (3) = 1 & ⟹ f (2) = 4, f (3) = 2 & ⟹ f (2) = 3, f (3) = 3 & ⟹ f (2) = 2 .

The admissible ordered pairs are:

(f (2), f (3)) = (4, 1), (3, 2), (2, 3) .

Each pair contains distinct values belonging to the set. The following pair satisfies the sum but fails the strict inequality:

(f (2), f (3)) = (1, 4), f (3) = 4 ⧸ < 4 .

Use this counting order: fix the linked pair, choose an unused allowed image for input 1, then arrange the remaining images. Starting with three choices for input 1 in every case ignores whether an output has already been used.

How do the three cases give exactly 48 functions?

The three cases contribute 12, 18 and 18 functions. The difference comes entirely from whether the linked pair has already used output 4. Once the first three inputs have distinct images, the remaining inputs, 4, 5 and 6, map bijectively to the three unused outputs.

There are no further restrictions on these remaining inputs. Their assignments therefore number: $3! = 3 \times 2 \times 1 = 6 .$

Case 1: the linked pair has used output 4. Only outputs 5 and 6 remain allowed for input 1.

(f (2), f (3)) & = (4, 1), f (1) & \in {5, 6}, Remaining assignments & = 3! = 6, Subtotal & = 2 \times 3! = 12 .

Choosing output 5 for input 1 leaves outputs 2, 3 and 6 for the remaining inputs. Choosing output 6 instead leaves outputs 2, 3 and 5. Each selection gives six assignments.

Case 2: the linked pair uses only outputs 3 and 2. All three allowed images for input 1 remain available.

(f (2), f (3)) & = (3, 2), f (1) & \in {4, 5, 6}, Remaining assignments & = 3! = 6, Subtotal & = 3 \times 3! = 18 .

Case 3: reversing the previous pair uses the same outputs. All three allowed images for input 1 remain available.

(f (2), f (3)) & = (2, 3), f (1) & \in {4, 5, 6}, Remaining assignments & = 3! = 6, Subtotal & = 3 \times 3! = 18 .

These cases are exhaustive and mutually exclusive: they cover all three possible images of input 3, and a function belongs to exactly one case. Add their totals, rather than multiplying them:

12 + 18 + 18 = (2 + 3 + 3) \times 6 = 48 .

The correct choice is option B.

Why does the calculation giving 72 fail?

The count of 72 loses two restrictions: the upper bound on the image of input 3 and the requirement that images stay distinct. Satisfying the sum condition is not enough. Every counted assignment must satisfy all conditions simultaneously.

The incorrect calculation is: $4 \times 3 \times 3! = 72 .$

Its factors represent four ordered positive pairs summing to 5, three presumed choices for the image of input 1, and six remaining assignments.

First error: the fourth pair is forbidden because its second entry violates the strict upper bound:

(f (2), f (3)) = (1, 4), f (3) = 4 ⧸ < 4 .

Second error: even among the three valid pairs, the following pair has already consumed output 4:

(f (2), f (3)) = (4, 1) .

Input 1 therefore has two choices, not three. Removing the forbidden pair but retaining a constant factor of three still gives an incorrect total: $3 \times 3 \times 3! = 54 .$

The multiplication principle is sound. The mistake is treating a case-dependent choice count as a constant.

Reusable rule: enumerate the linked variables, cross out used images, and only then multiply within each case.

What changes when you alter one restriction?

Tightening the first inequality gives 36; removing the second gives 60. Both questions below are original practice variants of the supplied problem, not additional verified PYQs. In each, check which allowed outputs remain after fixing the linked pair.

Question 1: Count one-one functions with these conditions:

A = {1, 2, 3, 4, 5, 6}, f : A \to A,

f (1) > 4, f (3) < 4, f (2) + f (3) = 5 .

The same three linked pairs remain valid:

(f (2), f (3)) = (4, 1), (3, 2), (2, 3) .

Neither output 5 nor output 6 appears in any pair. The image of input 1 therefore has exactly two choices in every case:

(2 + 2 + 2) \times 3! = 36 .

Lesson: a constant factor is legitimate when availability has actually been checked in every case.

Question 2: For the same set, count one-one functions satisfying only:

f : A \to A, f (1) > 3, f (2) + f (3) = 5 .

There is now no upper-bound restriction on the image of input 3. The ordered pairs and corresponding numbers of available images for input 1 are:

(f (2), f (3)) = (4, 1) & : 2, (f (2), f (3)) = (3, 2) & : 3, (f (2), f (3)) = (2, 3) & : 3, (f (2), f (3)) = (1, 4) & : 2 .

The total is:

(2 + 3 + 3 + 2) \times 3! = 60 .

Lesson: admitting the fourth pair still does not justify 72. Output 4 is unavailable to input 1 in both pairs containing it.

The answers 36 and 60 belong to these modified questions, not the original. Before multiplying in your next restricted-function problem, write the unused allowed outputs beside every linked pair.

Next step: photograph a doubt on JEEnius and photograph any question you are stuck on and get a step-by-step solution, with a free-body diagram when the question needs one (20 free).

If that step was the hard part, work through Intermolecular Forces JEE 2025: Why C and D Are Correct.

Frequently asked questions

How do you get 48 in this JEE permutations and combinations question?

For one-one functions from {1,2,3,4,5,6} to itself with f(1)>3, f(3)<4 and f(2)+f(3)=5, the valid ordered pairs (f(2),f(3)) are (4,1), (3,2) and (2,3). These leave 2, 3 and 3 choices respectively for f(1), followed by 3! assignments of the remaining outputs. The total is (2+3+3) × 6 = 48, option B.

Why is 72 wrong in the restricted-function question?

The calculation 4 × 3 × 3! includes the pair (f(2),f(3))=(1,4), which violates f(3)<4. It also assumes three choices for f(1) in every case, although the valid pair (4,1) leaves only 5 and 6 available. Removing only the forbidden pair gives 54; correcting both errors gives 48.

Why do we multiply by 3! after choosing the first three outputs?

Once f(1), f(2) and f(3) have distinct outputs, three unused outputs remain for inputs 4, 5 and 6. These inputs have no further restrictions, so the remaining outputs can be assigned bijectively in 3! = 6 ways.

When should I add cases instead of multiplying them in P&C?

Add case totals when the cases are mutually exclusive and together cover all valid possibilities. In this problem, f(3)=1, 2 and 3 define three such cases, contributing 12, 18 and 18 functions. Multiply choices within each case only after checking how earlier selections affect the available outputs.