Q.21. Let A = {1, 2, 3, 4, 5, 6}. The number of one-one functions f: A A such that f(1) 3, f(3) 4

Q: What is the correct answer for JEE Main Mathematics Q21?

The correct answer is 72. Sol. We are looking for one-one functions f: A $\to$ A, where A = {1, 2, 3, 4, 5, 6}. Conditions: 1. f(1) $\ge$ 3: f(1) can be 3, 4, 5, 6. 2. f(3) $\le$ 4: f(3) can be 1, 2, 3, 4. 3. f(2) + f(3) = 5. Since f is one-one, f(1), f(2), f(3) must be distinct. Let's analyze condition 3: f(2) + f(3) = 5. Possible pairs for (f(2), f(3)) such that f(2) $\ne$ f(3): - If f(2)=1, f(3)=4. (Valid, f(3)=4 $\le$

Q: Q.21. Let A = {1, 2, 3, 4, 5, 6}. The number of one-one functions f: A A such that f(1) 3, f(3) 4 and f(2) + f(3) = 5, is

The correct answer is 72. Sol. We are looking for one-one functions f: A $\to$ A, where A = {1, 2, 3, 4, 5, 6}. Conditions: 1. f(1) $\ge$ 3: f(1) can be 3, 4, 5, 6. 2. f(3) $\le$ 4: f(3) can be 1, 2, 3, 4. 3. f(2) + f(3) = 5. Since f is one-one, f(1), f(2), f(3) must be distinct. Let's analyze condition 3: f(2) + f(3) = 5. Possible pairs for (f(2), f(3)) such that f(2) $\ne$ f(3): - If f(2)=1, f(3)=4. (Valid, f(3)=4 $\le$

Question

Q.21. Let A = {1, 2, 3, 4, 5, 6}. The number of one-one functions f: A  A such that f(1)  3, f(3)  4 and f(2) + f(3) = 5, is

FirstInTest · Accepted Answer

Sol. We are looking for one-one functions f: A $	o$ A, where A = {1, 2, 3, 4, 5, 6}.
Conditions:
1. f(1) $\ge$ 3: f(1) can be 3, 4, 5, 6.
2. f(3) $\le$ 4: f(3) can be 1, 2, 3, 4.
3. f(2) + f(3) = 5.
Since f is one-one, f(1), f(2), f(3) must be distinct.
Let's analyze condition 3: f(2) + f(3) = 5.
Possible pairs for (f(2), f(3)) such that f(2) $
e$ f(3):
- If f(2)=1, f(3)=4. (Valid, f(3)=4 $\le$ 4)
- If f(2)=2, f(3)=3. (Valid, f(3)=3 $\le$ 4)
- If f(2)=3, f(3)=2. (Valid, f(3)=2 $\le$ 4)
- If f(2)=4, f(3)=1. (Valid, f(3)=1 $\le$ 4)
Case 1: (f(2), f(3)) = (1, 4)
   - f(1) $\ge$ 3. f(1) cannot be 1 or 4 (since f is one-one). So f(1) can be 3, 5, 6 (3 choices).
   - The remaining 3 elements in A ({1,2,3,4,5,6} - {f(1), f(2), f(3)}) can be mapped to the remaining 3 elements in A in 3! ways.
   - Number of functions for this case = 3 $	imes$ 3! = 3 $	imes$ 6 = 18.
Case 2: (f(2), f(3)) = (2, 3)
   - f(1) $\ge$ 3. f(1) cannot be 2 or 3. So f(1) can be 4, 5, 6 (3 choices).
   - Number of functions for this case = 3 $	imes$ 3! = 3 $	imes$ 6 = 18.
Case 3: (f(2), f(3)) = (3, 2)
   - f(1) $\ge$ 3. f(1) cannot be 3 or 2. So f(1) can be 4, 5, 6 (3 choices).
   - Number of functions for this case = 3 $	imes$ 3! = 3 $	imes$ 6 = 18.
Case 4: (f(2), f(3)) = (4, 1)
   - f(1) $\ge$ 3. f(1) cannot be 4 or 1. So f(1) can be 3, 5, 6 (3 choices).
   - Number of functions for this case = 3 $	imes$ 3! = 3 $	imes$ 6 = 18.
Total number of functions = 18 + 18 + 18 + 18 = 72.

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