Q10 of 194 JEE Main 2019 (11 Jan Shift 2) The number of functions f from {1,2,3,...,20} onto {1,2,3,...,20} such that f(k) is a multiple of 3, whenever k is a multiple of 4 is: A 6^5 x (15)! B 5! x 6! c (15)! x 6! D 5^6 x 15

✍️ Key Concepts To Crack The Problem

1. Onto Function with Equal Sizes ⇒ Bijection
   If a function $f\colon A \to B$ is onto (surjective) and $|A| = |B|$, then $f$ is automatically one-to-one as well.
   Hence the function is just a permutation of the 20 elements.

2. Restricted Positions
   The domain elements that are multiples of 4 are $\{4, 8, 12, 16, 20\}$ (5 of them).
   Their images must come from the set of multiples of 3 in the codomain: $\{3, 6, 9, 12, 15, 18\}$ (6 of them).

3. Counting a Bijection with Restrictions
   • Step 1: Choose an injective (no-repeats) assignment from the 5 restricted domain elements to the 6 allowed codomain elements.
   • Step 2: After removing those 5 chosen images, the remaining 15 domain elements must match up bijectively with the remaining 15 codomain elements.

4. Permutations Not Combinations
   Because the exact which-goes-where matters, we count ordered choices (permutations), not just subsets.

🧒 Imagine a Game of Matching Boxes and Toys

1. 20 Boxes & 20 Toys
   Think of 20 numbered boxes (1 to 20) and 20 different toys also numbered 1 to 20.

2. Special Rule
   The boxes numbered 4, 8, 12, 16, 20 are special.
   For these special boxes you are only allowed to put in a toy whose number is a multiple of 3 (3, 6, 9, 12, 15, 18).

3. Onto = Everyone Gets Chosen
   The word "onto" just means every toy must land in exactly one box.
   Because we have the same number of boxes and toys (20 each), this forces a perfect one-to-one matching (mathematicians call it a permutation).

4. How Many Ways?
   • First choose which 5 of the 6 special toys go into the 5 special boxes.
   • Then arrange the rest of the toys in the remaining 15 ordinary boxes any way you like.

That counting gives the final answer!

Step-by-Step Solution

1. Identify Special Domain Elements
   $S = \{4, 8, 12, 16, 20\} \quad (|S| = 5)$

2. Allowed Images for S
   $M = \{3, 6, 9, 12, 15, 18\} \quad (|M| = 6)$

3. Onto ⇒ Bijection
   Because domain and codomain both contain 20 elements, an onto function is a permutation of the set $\{1,2,\dots,20\}$.

4. Assign Images to Special Elements
   We need an injective mapping from 5 elements of $S$ to 6 elements of $M$.
   Number of ways:

   $P(6,5) = 6 \times 5 \times 4 \times 3 \times 2 = 6!$

5. Assign Remaining Images
   After fixing those 5 images, we have 15 domain elements left and 15 codomain elements left.
   They can be matched in $15!$ ways (a permutation of 15 items).

6. Total Count

   $\text{Total} = 6! \times 15!$

7. Choose the Correct Option
   Option C matches $15! \times 6!$.

Final Answer: (15)! × 6!