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JEE Mathematics 2026 Question with Solution

The number of strictly increasing functions ff from the set {1,2,3,4,5,6}\{1, 2, 3, 4, 5, 6\} to the set {1,2,3,…,9}\{1, 2, 3, \ldots, 9\} such that f(i)>if(i) > i for 1≤i≤61 \le i \le 6, is equal to:

  • A

    2222

  • B

    2727

  • C

    2121

  • D

    2828

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: We need the number of strictly increasing functions ff from {1,2,3,4,5,6}\{1,2,3,4,5,6\} to {1,2,…,9}\{1,2,\ldots,9\} such that f(i)>if(i)>i for 1≤i≤61 \le i \le 6.

Find: The total number of such functions.

Let

yi=f(i)y_i = f(i)

Then the conditions become

y1<y2<y3<y4<y5<y6y_1<y_2<y_3<y_4<y_5<y_6

and

yi>ifor all i=1,2,…,6y_i > i \quad \text{for all } i=1,2,\ldots,6

Since yiy_i are integers, yi>iy_i > i means

yi≥i+1y_i \ge i+1

Transformation to a combinations problem

Now define

ci=yi−ic_i = y_i - i

From yi≥i+1y_i \ge i+1, we get

ci≥1c_i \ge 1

Also, from the strictly increasing condition,

yi<yi+1y_i < y_{i+1}

so

ci+i<ci+1+(i+1)c_i + i < c_{i+1} + (i+1)

which gives

ci−1<ci+1c_i - 1 < c_{i+1}

Since the values are integers, this is equivalent to

ci≤ci+1c_i \le c_{i+1}

Hence

1≤c1≤c2≤c3≤c4≤c5≤c61 \le c_1 \le c_2 \le c_3 \le c_4 \le c_5 \le c_6

Because the codomain is {1,2,…,9}\{1,2,\ldots,9\}, we also have

y6≤9y_6 \le 9

Thus

c6+6≤9c_6 + 6 \le 9

so

c6≤3c_6 \le 3

Therefore we must count the number of non-decreasing sequences of length 66 formed from {1,2,3}\{1,2,3\}:

1≤c1≤c2≤c3≤c4≤c5≤c6≤31 \le c_1 \le c_2 \le c_3 \le c_4 \le c_5 \le c_6 \le 3

This is the same as choosing 66 items from 33 types with repetition allowed. Hence the number of such sequences is

(3+6−16)=(86)\binom{3+6-1}{6} = \binom{8}{6}

Now

(86)=(82)=8×72×1=28\binom{8}{6} = \binom{8}{2} = \frac{8 \times 7}{2 \times 1} = 28

Therefore, the number of such strictly increasing functions is 2828. The correct option is D.

Common mistakes

  • Treating a strictly increasing function as an arbitrary function. That is wrong because once the six image values are chosen in increasing order, the function is fixed. Instead, count only valid increasing sequences.

  • Using the condition f(i)>if(i)>i as if it were only needed for one or two values of ii. The inequality must hold for every i=1,2,…,6i=1,2,\ldots,6. After substitution, this gives the lower bound ci≥1c_i\ge 1 for all terms.

  • Missing the upper bound from the codomain. Since f(6)≤9f(6)\le 9, we must have c6≤3c_6\le 3. Without this restriction, the count becomes too large.

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