NIMCET 2026 — Mathematics PYQ

To solve this problem, we need to count the number of injective functions $f$ that map even numbers in the domain to even numbers in the codomain.

Step 1: Identify the elements

• Domain: $\{1,2,3,4\}$

  • Even numbers in Domain ($D_{even}$): $\{2,4\}$ (Total: $2$)

  • Odd numbers in Domain ($D_{odd}$): $\{1,3\}$ (Total: $2$)

• Codomain: $\{1,2,3,4,5,6,7,8\}$

  • Even numbers in Codomain ($C_{even}$): $\{2,4,6,8\}$ (Total: $4$)

  • Odd numbers in Codomain ($C_{odd}$): $\{1,3,5,7\}$ (Total: $4$)

Step 2: Calculate mappings

• Mapping even numbers: We must map the $2$ even numbers from the domain to the $4$ even numbers in the codomain. Since the function must be injective, the number of ways is given by the permutation $P(4,2)$:

  $$P(4,2)=4\times 3=12$$

• Mapping odd numbers: We have $2$ odd numbers remaining in the domain. We have already used $2$ numbers from the codomain for the even mappings. Therefore, the remaining number of available elements in the codomain is $8-2=6$. The number of ways to map the $2$ odd numbers into the $6$ available slots is given by $P(6,2)$:

  $$P(6,2)=6\times 5=30$$

• Total injective functions ($n$):

  $$n=12\times 30=360$$

Step 3: Prime Factorization

Now, we express $n$ in the form $2^a{3}^b{5}^c$:

$$360=36\times 10$$

$$360=(2^2\times 3^2)\times (2^1\times 5^1)$$

$$360=2^{2+1}\times 3^2\times 5^1$$

$$n=2^3\times 3^2\times 5^1$$

By comparing $n=2^a{3}^b{5}^c$ with $2^3\times 3^2\times 5^1$, we get:

• $a=3$

• $b=2$

• $c=1$

Final Step: Calculate $a+b+c$

$$a+b+c=3+2+1=6$$

Correct Option: 2. 6