The number of strictly increasing functions f from the set {1, 2, 3, 4, 5, 6} to the set {1, 2, 3, .

Q: Wait, why can't we just use the inclusion-exclusion principle to subtract the cases where $f(i) = i$?

That is a great intuition! However, inclusion-exclusion becomes incredibly messy here because the condition $f(i) = i$ is coupled with the strictly increasing requirement. The transformation $h(i) = f(i) - i$ is a much more elegant 'JEE-style' shortcut that turns a constrained problem into a standard stars-and-bars counting problem.

Q: Why is $h(i)$ non-decreasing? I am struggling to see the connection to $f(i+1) > f(i)$.

Excellent question. Since $f$ is strictly increasing, $f(i+1) \ge f(i) + 1$. If we subtract $i+1$ from both sides, we get $f(i+1) - (i+1) \ge f(i) - i$. By our definition, this is exactly $h(i+1) \ge h(i)$. It is a beautiful algebraic consequence of the strictly increasing property!

Q: How do we know the upper bound for $h(6)$ is 3?

Look at the codomain: the maximum value $f(6)$ can take is 9. Since $h(6) = f(6) - 6$, the maximum value for $h(6)$ is $9 - 6 = 3$. Since $h(i)$ is non-decreasing, all $h(i)$ values are bounded by this maximum of 3.

Question

The number of strictly increasing functions $f$ from the set $\{1, 2, 3, 4, 5, 6\}$ to the set $\{1, 2, 3, \dots, 9\}$ such that $f(i) 
eq i$ for $1 \le i \le 6$, is equal to :

Options:

(A) 27

(B) 22

(C) 21

(D) 28

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE mastery! Today, we are going to unravel a problem that, at first glance, seems like a simple counting exercise but is actually a masterclass in mathematical transformation.

We are tasked with finding the number of strictly increasing functions $f$ from the set $A = \{1, 2, 3, 4, 5, 6\}$ to the set $B = \{1, 2, 3, \dots, 9\}$ such that $f(i) 
eq i$ for all $i \in A$.

The 'strictly increasing' rule implies $f(1) < f(2) < f(3) < f(4) < f(5) < f(6)$. This means that once we choose six distinct values from the codomain, there is only one way to arrange them to satisfy the increasing condition.

### The Forbidden Zone

Let us visualize the constraint. For any strictly increasing function on integers, the minimum value an element can take is its index, so $f(i) \ge i$.

The problem explicitly states that $f(i) 
eq i$. Since $f(i)$ must be at least $i$ and cannot be $i$, it must satisfy the condition $f(i) \ge i+1$.

This shift in the lower bound is the key to unlocking the problem. We have effectively moved from a constrained set to a more manageable range.

### The Elegant Transformation

Let us define a new function $h(i) = f(i) - i$. If $f(i) \ge i+1$, then $f(i) - i \ge 1$, which implies $h(i) \ge 1$.

Regarding monotonicity, since $f$ is strictly increasing, we know $f(i+1) \ge f(i) + 1$. Subtracting $i+1$ from both sides yields:

$$f(i+1) - (i+1) \ge f(i) - i \Rightarrow h(i+1) \ge h(i)$$

Our new function $h$ is non-decreasing. We have transformed the problem into finding the number of sequences satisfying:

$$1 \le h(1) \le h(2) \le h(3) \le h(4) \le h(5) \le h(6)$$

Since the maximum value in our codomain is 9, we have $f(6) \le 9$. Consequently, $h(6) = f(6) - 6 \le 9 - 6 = 3$. Our sequence is bounded by:

$$1 \le h(1) \le h(2) \le h(3) \le h(4) \le h(5) \le h(6) \le 3$$

### Final Calculation

We need to choose 6 values for $h(i)$ from the set $\{1, 2, 3\}$ such that they are non-decreasing. This is a classic combinations with repetition problem.

We are choosing $r=6$ items from $n=3$ types. The formula for combinations with repetition is given by $\binom{n+r-1}{r}$. Substituting our values:

$$\binom{3+6-1}{6} = \binom{8}{6}$$

Using the symmetry property $\binom{n}{r} = \binom{n}{n-r}$, we calculate:

$$\binom{8}{6} = \binom{8}{2} = \frac{8 	imes 7}{2 	imes 1} = 28$$

The total number of functions that satisfy the given conditions is **28**. By transforming the problem, we stripped away the complexity to reveal the elegant structure underneath.

Animated Solution for Mathematics - Permutations and Combinations: The number of strictly increasing functions $f$ from the set ${1, 2, 3, 4, 5, 6}$ to the set ${1, 2, 3, \dots, 9}$ such that $f (i) \neq = i$ for $1 \leq i \leq 6$ , is equal to :

Select Answer:

Visualized Solution

Understanding the Sets

Strictly Increasing Property

Applying the Forbidden Mapping

Introducing the Transformation $h (i)$

Finding the Upper Bound

Analyzing Monotonicity of $h (i)$

The Complete Sequence

Combinations with Repetition

Substituting the Values

Final Calculation

Conclusion

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Forbidden Zone

The Elegant Transformation

Final Calculation

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Select Answer:

Visualized Solution

Understanding the Sets

Strictly Increasing Property

Applying the Forbidden Mapping

Introducing the Transformation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h(i)

Finding the Upper Bound

Analyzing Monotonicity of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h(i)

The Complete Sequence

Combinations with Repetition

Substituting the Values

Final Calculation

Conclusion

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Forbidden Zone

The Elegant Transformation

Final Calculation

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