Consider the sets A={(x, y) ∈ ℝ × ℝ: x²+y²=25}, B={(x, y) ∈ ℝ × ℝ: x²+9 y²=144}, C={(x, y) ∈ ℤ × ℤ: x²+y² ≤ 4} and D=A ∩ B. The total number of one-one functions from the set D to the set C is:

Solution

Maths sir · 4 steps

First count how many points sit in D and in C; then a one-one map is just an ordered pick of images.

1. Find D. Points of D lie on both the circle and the ellipse. Subtract x² + y² = 25 from x² + 9y² = 144 to get 8y² = 119, so y² = 119/8 and x² = 25 - 119/8 = 81/8.
2. Four points. Both x² and y² are positive, so x = ±9/2√2 and y = ±√(119/8). Every sign pair works, so D has 4 elements.

   Value of x	Allowed y (integers, y² ≤ 4 - x²)	Points
   0	-2, -1, 0, 1, 2	5
   ± 1	-1, 0, 1	3 + 3 = 6
   ± 2	0	1 + 1 = 2

3. Count C. Adding the rows gives 5 + 6 + 2 = 13 integer points in C.

   ¹³P₄ = 13 × 12 × 11 × 10 = 17160

4. So. The 4 points of D must go to 4 different points of C, chosen in order. That gives 17160 one-one functions, option (c).

The idea to keep. The number of one-one functions from a set of m elements to a set of n elements is ⁿPₘ. The real work is counting m and n, so list lattice points row by row.

Answer from the NTA final answer key. Solution by PrepWiser.