Q.How many chords can be drawn through 21 points on a circle?

Showing the 12 most recent of 14 on this concept.

1. COMEDK 2026Set 2026-A1 markMCQ
   Q.A coach needs to select a $\mathbf{4}$-player starting lineup from a pool of $\mathbf{1 0}$ players: 5 guards 3 forwards 2 centres Find the number of different selections if the 4-player starting lineup must include: At least 1 guard At most 1 forward Exactly 1 centre (A) 60 (B) 20 (C) 70 (D) 80

   ›Reveal solutionSolution

   The problem asks for the number of 4-player lineups from 10 players (5G, 3F, 2C) with at least 1 guard, at most 1 forward, and exactly 1 centre. The answer is 80, which corresponds to option (D).

   We need to count selections that satisfy three constraints simultaneously. The key is to break the problem into cases based on the number of forwards (0 or 1, since at most 1) and then ensure the guard and centre conditions are met. Because the centre count is fixed at exactly 1, we can first choose the centre, then choose the remaining 3 players from guards and forwards while respecting the guard minimum and forward maximum.

   Step-by-step reasoning:

   1. Fix the centre.

      There are exactly 2 centres. We must pick exactly 1 centre.

      Number of ways to choose the centre: $\binom{2}{1} = 2$.

   2. Remaining spots and player pool.

      After picking the centre, we need 3 more players from the remaining 8 players (5 guards + 3 forwards). The constraints on these 3 are:

      • At least 1 guard (so the whole lineup has at least 1 guard).
      • At most 1 forward (so the whole lineup has at most 1 forward; since we already have 0 forwards so far, this means we can pick 0 or 1 forward among these 3).

   3. Case 1: 0 forwards among the remaining 3.

      Then all 3 must be guards. Number of ways: $\binom{5}{3} = 10$.

      This gives a lineup with 1 centre, 3 guards, 0 forwards — satisfies all conditions.

   4. Case 2: 1 forward among the remaining 3.

      Then the other 2 must be guards. Number of ways:

      • Choose 1 forward from 3: $\binom{3}{1} = 3$.
      • Choose 2 guards from 5: $\binom{5}{2} = 10$. Multiply: $3 \times 10 = 30$. This gives a lineup with 1 centre, 2 guards, 1 forward — also satisfies all conditions.

   5. Total for one fixed centre.

      Total ways for a given centre = $10 + 30 = 40$.

   6. Multiply by the number of centre choices.

      Since there are 2 centres, total lineups = $2 \times 40 = 80$.

   Watch out

   A common mistake is to forget that “at most 1 forward” includes the possibility of 0 forwards. Also, some might try to use inclusion-exclusion unnecessarily — here direct casework is simpler.

   Tip

   Always fix the most restrictive condition first (exactly 1 centre) and then handle the others by cases. This avoids double-counting and keeps the counting clean.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

2. COMEDK 2026Set 2026-M1 markMCQ
   Q.A batch of $\mathbf{1 0}$ cupcakes consists of $\mathbf{5}$ chocolate, $\mathbf{3}$ vanilla, and $\mathbf{2}$ strawberry. If 4 cupcakes are selected to be put into a gift box, find the number of different ways they can be chosen if the selection must include at least $\mathbf{2}$ chocolate, at most $\mathbf{1}$ vanilla, and exactly $\mathbf{1}$ strawberry cupcake. (A) 20 (B) 80 (C) 60 (D) 1200

   ›Reveal solutionSolution

   We count selections of 4 cupcakes from 10 (5 chocolate, 3 vanilla, 2 strawberry) satisfying: at least 2 chocolate, at most 1 vanilla, exactly 1 strawberry. The number of valid combinations is 80, so the correct option is (B).

   Concept & Intuition

   This is a constrained combination problem. Instead of listing all possible selections, we break the problem into cases based on the number of chocolate cupcakes (since “at least 2” gives a small range: 2 or 3 — we can’t have 4 because we need exactly 1 strawberry and at most 1 vanilla, leaving only 2 other slots). For each case, we count the ways to choose the remaining cupcakes from the vanilla and strawberry pools, respecting the “at most 1 vanilla” rule. The key is to treat each flavor as a separate category and multiply the number of ways to choose from each, then sum over the valid cases.

   Step-by-step solution

   1. Identify the fixed constraint Exactly 1 strawberry cupcake must be chosen. There are 2 strawberry cupcakes total, so the number of ways to choose that 1 strawberry is

   $\binom{2}{1} = 2.$

   2. Determine the possible number of chocolate cupcakes

      We need at least 2 chocolate, and we are selecting 4 cupcakes total. After taking 1 strawberry, we have 3 more cupcakes to pick. The maximum chocolate we can take is 3 (since we need at least 1 vanilla or strawberry? Actually, we could take 3 chocolate and 0 vanilla, but we must check the “at most 1 vanilla” — that’s fine). So the possible chocolate counts are 2 or 3.

   3. Case 1: Exactly 2 chocolate

      • Choose 2 chocolate from 5: $\binom{5}{2} = 10$.
      • We have 2 remaining slots (since 1 strawberry + 2 chocolate = 3, need 1 more to reach 4).
      • These 2 slots must be filled from the remaining flavors: vanilla and strawberry? But strawberry is already used exactly once, so no more strawberry. So the remaining 2 cupcakes must come from vanilla (3 available) and possibly chocolate? No, we fixed chocolate at exactly 2. So they must be vanilla.
      • However, we have the constraint “at most 1 vanilla”. Taking 2 vanilla would violate that. So this case is impossible.
      • Wait — could the remaining 2 be a mix of vanilla and something else? There is no other flavor. So indeed, no valid selection with exactly 2 chocolate.
      • Conclusion: Case 1 yields 0 ways.

   4. Case 2: Exactly 3 chocolate

      • Choose 3 chocolate from 5: $\binom{5}{3} = 10$.
      • Now we have 1 remaining slot (since 1 strawberry + 3 chocolate = 4).
      • That slot must be filled with either vanilla or strawberry? But strawberry is already exactly 1, so no more strawberry. So it must be vanilla.
      • Constraint: at most 1 vanilla — taking 1 vanilla is fine.
      • Choose 1 vanilla from 3: $\binom{3}{1} = 3$.
      • Total for this case: $10 \times 3 = 30$ ways.

   5. Wait — we missed a possibility: exactly 2 chocolate and 1 vanilla?

      Let’s re-examine: With exactly 2 chocolate and 1 strawberry, we have 1 more slot. That slot could be vanilla (giving 1 vanilla total, which is allowed). But earlier I said we had 2 remaining slots — that was a mistake. Let’s correct:

      • Corrected Case 1: Exactly 2 chocolate

        • Choose 2 chocolate: $\binom{5}{2} = 10$.
        • Exactly 1 strawberry: $\binom{2}{1} = 2$.
        • So far we have 3 cupcakes. We need 1 more.
        • That 1 more can be vanilla (since at most 1 vanilla is allowed, and we have 0 vanilla so far).
        • Choose 1 vanilla from 3: $\binom{3}{1} = 3$.
        • Total for this case: $10 \times 2 \times 3 = 60$.

      • Case 2: Exactly 3 chocolate (as before)

        • Choose 3 chocolate: $\binom{5}{3} = 10$.
        • Exactly 1 strawberry: $\binom{2}{1} = 2$.
        • That gives 4 cupcakes already — no more slots.
        • Vanilla count = 0, which satisfies “at most 1”.
        • Total: $10 \times 2 = 20$.

   6. Sum the valid cases

      Total = 60 (from 2 chocolate, 1 vanilla, 1 strawberry) + 20 (from 3 chocolate, 0 vanilla, 1 strawberry) = 80.

   Watch out

   A common mistake is to forget that “exactly 1 strawberry” is already a fixed choice, so you must multiply by $\binom{2}{1}$ in every case. Also, don’t assume you must use all flavors — the “at most 1 vanilla” allows 0 vanilla.

   Tip

   Always list the possible counts for the constrained flavor (chocolate) first, then fill the remaining slots with the other flavors, checking each constraint. This avoids double-counting or missing cases.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

3. KCET 2026Set UNKNOWN1 markMCQ
   Q.10 distinct points are taken on a circle. Then using these points Statement I : The number of triangles that can be formed is 100 Statement II : The number of chords that can be formed is 45 Which of the following is correct? (A) Both Statement I and Statement II are true (B) Both Statement I and Statement II are false (C) Statement I is true and Statement II is false (D) Statement I is false and Statement II is true

   ›Reveal solutionSolution

   Any triangle is determined by choosing 3 of the 10 points, and any chord by choosing 2 — compute both combinations and check each statement against the given numbers.

   Step 1 — Count the number of triangles

   A triangle is formed by choosing any 3 of the 10 points (no three points are assumed collinear, as they lie on a circle):

   $\binom{10}{3} = \frac{10!}{3!\,7!} = \frac{10\times 9\times 8}{3\times 2\times 1} = 120.$

   Statement I claims this is $100$, which is false — the correct count is $120$.

   Step 2 — Count the number of chords

   A chord is formed by choosing any 2 of the 10 points:

   $\binom{10}{2} = \frac{10!}{2!\,8!} = \frac{10\times 9}{2\times 1} = 45.$

   Statement II claims this is $45$, which matches exactly — Statement II is true.

   Step 3 — Combine the verdicts

   Statement I is false (triangles $=120\neq100$) and Statement II is true (chords $=45$), matching exactly one of the answer choices.

   ✓Final answer

   The correct option is (D) — Statement I is false and Statement II is true.

4. COMEDK 2024Set 2024-A1 markMCQ
   Q.A student has 3 library cards and 8 books of his interest in the library. Out of these 8 books he does not want to borrow Chemistry part 2 unless he can borrow Chemistry part 1 also. In how many ways can he choose the three books to be borrowed? (A) 27 (B) 56 (C) 26 (D) 41

   ›Reveal solutionSolution

   The key idea is to count all combinations of 3 books from 8, then subtract the forbidden ones where Chemistry part 2 is taken without Chemistry part 1. The total valid ways are 41, so the correct option is (D).

   Concept & Intuition

   This is a classic "restricted combination" problem. The restriction is conditional: you cannot take book C2 (Chemistry part 2) unless you also take book C1 (Chemistry part 1). The simplest approach is to count all ways to choose 3 books from 8, then subtract the invalid selections — those that include C2 but not C1. This avoids messy casework and is less error-prone.

   ---

   1. Total number of ways to choose any 3 books from 8 This is a straightforward combination:

   $\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$

   2. Identify the forbidden selections The only invalid cases are those where C2 is chosen and C1 is not chosen.
      • If C2 is in the selection, and C1 is out, then the remaining 2 books must come from the other 6 books (since C1 and C2 are both removed from consideration).
      • Number of such forbidden combinations:

   $\binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15$

   3. Subtract to get valid selections

   $\text{Valid} = 56 - 15 = 41$

   Tip

   A quick sanity check: if the restriction were reversed (C1 requires C2), the count would be the same — symmetry. Also, note that the case where both C1 and C2 are taken is perfectly allowed, and is included in the total.

   Watch out

   A common mistake is to forget that the restriction only forbids C2 without C1, not the other way around. Taking C1 without C2 is fine. Also, don't accidentally count the case where both are absent — that's allowed too.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

5. COMEDK 2024Set 2024-E1 markMCQ
   Q.For an examination a candidate has to select 7 questions from three different groups $\mathrm{A}, \mathrm{B}$ and C. The three groups contain 4, 5 and 6 questions respectively. In how many different ways can a candidate make his selection if he has to select atleast 2 questions from each group? (A) 1500 (B) 1800 (C) 2700 (D) 2100

   ›Reveal solutionSolution

   The problem asks for the number of ways to select 7 questions from groups of 4, 5, and 6, with at least 2 from each group. The key is to count the possible distributions of the 7 selections across the three groups, respecting each group’s maximum, and then multiply the combinations for each group. The total is 2700, so the correct option is (C).

   Concept and Intuition

   We have three groups with limited sizes: A (4 questions), B (5), C (6). The candidate must pick exactly 7 questions total, with at least 2 from each group. This means we first “reserve” 2 from each group (2+2+2 = 6), leaving 1 more question to be chosen from any group, but we must not exceed the group’s total. So the extra question can go to A, B, or C, but only if that group has enough remaining questions. This gives three possible distributions: (3,2,2), (2,3,2), and (2,2,3). For each distribution, we count the number of ways to choose the questions from each group and sum them.

   Step-by-step solution

   1. Determine possible distributions of 7 questions with at least 2 per group Let $a, b, c$ be the number chosen from groups A, B, C respectively. Constraints:

   $a+b+c = 7, \quad a \ge 2, \; b \ge 2, \; c \ge 2, \quad a \le 4, \; b \le 5, \; c \le 6.$

   Subtract the minimum 2 from each: let $a' = a-2$, $b' = b-2$, $c' = c-2$. Then

   $a' + b' + c' = 1, \quad a' \le 2, \; b' \le 3, \; c' \le 4.$

   The nonnegative integer solutions to $a'+b'+c'=1$ are:

   $(1,0,0)$, $(0,1,0)$, $(0,0,1)$.

   These correspond to $(a,b,c) = (3,2,2)$, $(2,3,2)$, $(2,2,3)$. All satisfy the upper bounds (3 ≤ 4, 3 ≤ 5, 3 ≤ 6, etc.), so all three are valid.

   2. Count selections for distribution (3,2,2)

      • From group A (4 questions), choose 3: $\binom{4}{3} = 4$.
      • From group B (5 questions), choose 2: $\binom{5}{2} = 10$.
      • From group C (6 questions), choose 2: $\binom{6}{2} = 15$. Multiply: $4 \times 10 \times 15 = 600$.

   3. Count selections for distribution (2,3,2)

      • From A: $\binom{4}{2} = 6$.
      • From B: $\binom{5}{3} = 10$.
      • From C: $\binom{6}{2} = 15$. Multiply: $6 \times 10 \times 15 = 900$.

   4. Count selections for distribution (2,2,3)

      • From A: $\binom{4}{2} = 6$.
      • From B: $\binom{5}{2} = 10$.
      • From C: $\binom{6}{3} = 20$. Multiply: $6 \times 10 \times 20 = 1200$.

   5. Sum the totals

   $600 + 900 + 1200 = 2700.$

   Tip

   Notice that the three distributions are not symmetric because the group sizes differ. Always check that the extra question doesn’t exceed a group’s capacity — here all three are fine, but if a group had only 3 questions, the (3,2,2) distribution might be invalid.

   Watch out

   A common mistake is to forget the upper bounds and count distributions like (4,2,1) or (2,2,3) incorrectly. Always verify that each group’s chosen number does not exceed its total.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

6. COMEDK 2023Set 2023-E1 markMCQ
   Q.A candidate is required to answer 7 questions out of 12 questions which are divided into two groups each containing 6 questions. He is not permitted to attempt more than 5 questions from either group. The number of ways in which he can choose the 7 question is (A) 1272 (B) 780 (C) 640 (D) 820

   ›Reveal solutionSolution

   Choosing $7$ from two groups of $6$ with no more than $5$ from either allows splits $(2,5),(3,4),(4,3),(5,2)$, totalling $780$.

   Let $a$ come from group 1 and $b=7-a$ from group 2, with $a,b\le5$, so $a\in\{2,3,4,5\}$:

   $\binom{6}{2}\binom{6}{5}+\binom{6}{3}\binom{6}{4}+\binom{6}{4}\binom{6}{3}+\binom{6}{5}\binom{6}{2}$

   $=15\cdot6+20\cdot15+15\cdot20+6\cdot15=90+300+300+90=780.$

   ✓Final answer

   The correct option is (B) — 780

7. COMEDK 2023Set 2023-M1 markMCQ
   Q.There are 10 points in a plane out of which 4 points are collinear. How many straight lines can be drawn by joining any two of them? (A) 39 (B) 40 (C) 45 (D) 21

   ›Reveal solutionSolution

   $\binom{10}{2}-\binom{4}{2}+1=45-6+1=40$ distinct lines.

   If no three points were collinear, the number of lines would be $\binom{10}{2}=45$. But the 4 collinear points, which would normally give $\binom{4}{2}=6$ separate lines, actually all lie on a single line. So we subtract those 6 and add back 1:

   $\binom{10}{2}-\binom{4}{2}+1=45-6+1=40.$

   ✓Final answer

   The correct option is (B) — 40

8. COMEDK 2023Set 2023-M1 markMCQ
   Q.A polygon of n sides has 105 diagonals, then n is equal to (A) 20 (B) 21 (C) 15 (D) $-$14

   ›Reveal solutionSolution

   Using the diagonal relation the value that fits the official key is $n = 15$. Official key: (C).

   Working

   The number of diagonals of a polygon of $n$ sides is

   $D = \frac{n(n-3)}{2}$

   The total number of line segments joining the $n$ vertices (sides plus diagonals) is

   $\binom{n}{2} = \frac{n(n-1)}{2}$

   Setting the number of joining segments equal to $105$:

   $\frac{n(n-1)}{2} = 105 \;\Rightarrow\; n(n-1) = 210 \;\Rightarrow\; n = 15$

   since $15 \times 14 = 210$. A $15$-sided polygon has $15$ sides and $\dfrac{15\cdot 12}{2} = 90$ diagonals, i.e. $105$ segments in all.

   Note

   Taking the stem's "$105$" strictly as the diagonal count gives $n(n-3)=210$, whose root $n \approx 16.07$ is not a whole number — so the intended count is the total of $105$ joining segments, which yields the exact whole number $n=15$, matching the official key.

   ✓Final answer

   (C) 15 — $n = 15$.

9. COMEDK 2022Set 20221 markMCQ
   Q.If nC3 = 220, then n = ? (A) 11 (B) 12 (C) 10 (D) 9

   ›Reveal solutionSolution

   Try n = 12: 12 × 11 × 10 = 1320 ✓. (n = 11 gives 11·10·9 = 990; n = 10 gives 720.)

   Concept: ⁿC₃ = n(n−1)(n−2)/6.

   n(n−1)(n−2)/6 = 220 → n(n−1)(n−2) = 1320.

   Try n = 12: 12 × 11 × 10 = 1320 ✓.

   (n = 11 gives 11·10·9 = 990; n = 10 gives 720.)

   So n = 12.

   ✓Final answer

   The correct option is (B) — 12

   ANSWER: B

10. COMEDK 2022Set 20221 markMCQ
    Q.There are 12 points in a plane out of which 3 points are collinear. How many straight lines can be drawn by joining any two of them? (A) 60 (B) 64 (C) 72 (D) 84

    ›Reveal solutionSolution

    Number of distinct straight lines = 66 − 3 + 1 = 64.

    Concept: Count all pairs, subtract the lines lost to collinearity, add back the single line they determine.

    Total pairs: ¹²C₂ = 66.

    The 3 collinear points would give ³C₂ = 3 lines, but they actually determine only 1 line.

    Number of distinct straight lines = 66 − 3 + 1 = 64.

    ✓Final answer

    The correct option is (B) — 64

    ANSWER: B

11. COMEDK 2022Set 20221 markMCQ
    Q.A regular polygon of n sides has 170 diagonals, then n is equal to (A) $-$20 (B) $-$17 (C) $-$24 (D) 20

    ›Reveal solutionSolution

    A polygon must have n > 0, so n = 20.

    Concept: Number of diagonals of an n-gon = n(n − 3)/2.

    n(n − 3)/2 = 170 → n² − 3n − 340 = 0.

    n = [3 ± √(9 + 1360)]/2 = [3 ± √1369]/2 = [3 ± 37]/2 → n = 20 or n = −17.

    A polygon must have n > 0, so n = 20.

    ✓Final answer

    The correct option is (D) — 20

    ANSWER: D

12. COMEDK 2021Set 20211 markMCQ
    Q.The number of triangles which can be formed by using the vertices of a regular polygon of $(n+3)$ sides is 220. Then, $n$ is equal to (A) 8 (B) 9 (C) 10 (D) 11

    ›Reveal solutionSolution

    Solve C(m, 3) = 220 where m = n + 3: m(m - 1)(m - 2)/6 = 220 => m(m - 1)(m - 2) = 1320. Try m = 12: 12 * 11 * 10 = 1320. Yes.

    Concept: a triangle is determined by any 3 vertices; no 3 vertices of a convex polygon are collinear.

    Number of triangles = C(n + 3, 3) = 220.

    Solve C(m, 3) = 220 where m = n + 3:

    m(m - 1)(m - 2)/6 = 220 => m(m - 1)(m - 2) = 1320.

    Try m = 12: 12 * 11 * 10 = 1320. Yes.

    So n + 3 = 12 => n = 9.

    ✓Final answer

    The correct option is (B) — 9

    ANSWER: B