Question

Let U = {10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20}. The sides a, b, c of a triangle are selected from U, then which of the following statement(s) are INCORRECT

• A. Number of triangles that can be made = 364
• B. Number of triangles that can be made = 285

Answer 1

Answer: (B)

a

Answer 2

The set of possible side lengths is U = {10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20}. The size of the set is n = 11. We are selecting three sides a, b, c from U. Since the order of the sides does not matter for forming a triangle, and the sides can be equal, we are selecting a multiset of size 3 from U. Let the selected sides be x, y, z such that 10 ≤ x ≤ y ≤ z ≤ 20. The condition for these lengths to form a triangle is the triangle inequality: x + y > z. (The other two inequalities x + z > y and y + z > x are automatically satisfied since x, y, z are positive and x ≤ y ≤ z).

The total number of ways to select a multiset of size 3 from a set of size 11 is given by the formula for combinations with repetition:

Total number of multisets = $\binom{n+k-1}{k}$, where n=11 (size of U) and k=3 (number of sides).

Total number of multisets = $\binom{11+3-1}{3} = \binom{13}{3} = \frac{13 \times 12 \times 11}{3 \times 2 \times 1} = 13 \times 2 \times 11 = 286$.

This is the total number of ways to choose three side lengths (allowing repetition, order doesn't matter) from the set U.

Now, we need to find the number of these multisets that do NOT form a triangle. These are the multisets {x, y, z} with 10 ≤ x ≤ y ≤ z ≤ 20 such that x + y ≤ z. Since x ∈ U, x ≥ 10. Since y ∈ U, y ≥ 10. So, x + y ≥ 10 + 10 = 20. Since z ∈ U, z ≤ 20. The condition x + y ≤ z combined with x + y ≥ 20 and z ≤ 20 implies that x + y must be exactly 20 and z must be exactly 20. So, we are looking for multisets {x, y, z} from U such that 10 ≤ x ≤ y ≤ z = 20 and x + y = 20.

We need to find pairs (x, y) such that 10 ≤ x ≤ y ≤ 20 and x + y = 20. Let's check possible values for x starting from 10:

If x = 10, then 10 + y = 20, which gives y = 10. We check if this satisfies 10 ≤ x ≤ y ≤ 20: 10 ≤ 10 ≤ 10 ≤ 20. Yes, this is satisfied. The triplet (x, y, z) is (10, 10, 20).

If x = 11, then 11 + y = 20, which gives y = 9. We check if this satisfies 10 ≤ x ≤ y ≤ 20: 10 ≤ 11 ≤ 9 ≤ 20. This is not satisfied because 11 ≤ 9 is false.

If x > 10, then x ≥ 11. Since y ≥ x, y ≥ 11. Then x + y ≥ 11 + 11 = 22. This contradicts the condition x + y = 20.

So, the only pair (x, y) satisfying 10 ≤ x ≤ y and x + y = 20 is (10, 10).

This gives only one triplet (x, y, z) with 10 ≤ x ≤ y ≤ z ≤ 20 that satisfies x + y ≤ z, which is (10, 10, 20). In this case, x + y = z (20 = 20), which does not satisfy the strict inequality x + y > z required for a triangle. So, there is only 1 multiset ({10, 10, 20}) from U that does not form a triangle.

The number of triangles that can be made is the total number of multisets minus the number of multisets that do not form a triangle.

Number of triangles = 286 - 1 = 285.

Now let's check the given statements:

Statement a: Number of triangles that can be made = 364. This is INCORRECT.

Statement b: Number of triangles that can be made = 285. This is CORRECT.

The question asks for the INCORRECT statement(s). The INCORRECT statement is a.