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

Showing the 12 most recent of 31 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.A candidate is required to answer seven out of 12 questions which are divided into two parts, each containing 6 questions. If he is not permitted to answer more than 5 questions from each part, the different number of ways he can choose the 7 questions is (A) $820$ (B) $780$ (C) $720$ (D) $640$

   ›Reveal solutionSolution

   The "not more than 5 from either part" restriction only rules out the extreme splits (all 7 from one side is impossible anyway, but 6-1 and 1-6 are ruled out by the cap); summing the valid splits gives $780$.

   Concept and Intuition

   When a selection is split across two groups with an upper limit per group, list every valid way the total count can be divided between the groups, compute each split's count via the multiplication principle, and add them all up.

   Step-by-Step Solution

   1. Let $k$ = number of questions chosen from Part I, so $7-k$ come from Part II.
   2. Each part has only 6 questions, and no more than 5 may be taken from either part: so $k\le5$ and $7-k\le5 \Rightarrow k\ge2$. Thus $k\in\{2,3,4,5\}$.
   3. Number of ways for a given $k$ is $\binom{6}{k}\binom{6}{7-k}$.
   4. $k=2$: $\binom{6}{2}\binom{6}{5}=15\times6=90$.
   5. $k=3$: $\binom{6}{3}\binom{6}{4}=20\times15=300$.
   6. $k=4$: $\binom{6}{4}\binom{6}{3}=15\times20=300$.
   7. $k=5$: $\binom{6}{5}\binom{6}{2}=6\times15=90$.
   8. Total $=90+300+300+90=780$.

   Common Mistakes

   • Including $k=6$ or $k=1$, forgetting these violate the "not more than 5 from each part" restriction.
   • Using $\binom{12}{7}$ directly, which ignores the part-wise cap entirely.

   ✓Final answer

   The correct option is (B) — $780$.

   ANSWER: B

2. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.There are 10 cards numbered 1 to 10. The number of ways in which at least 3 cards can be chosen from these 10 cards is (A) $1023$ (B) $1013$ (C) $1024$ (D) $968$

   ›Reveal solutionSolution

   Count all subsets of the 10 cards ($2^{10}$) and subtract the subsets of size 0, 1, and 2 to leave only "at least 3" selections.

   Concept and Intuition

   "At least 3" is most easily handled by the complement: total ways to choose any subset of a 10-element set, minus the ways to choose a subset smaller than 3 (i.e. size 0, 1, or 2).

   Step-by-Step Solution

   1. Total number of subsets of 10 distinct cards $= 2^{10} = 1024$ (each card is either chosen or not).
   2. Number of ways to choose exactly 0 cards: $\binom{10}{0}=1$.
   3. Number of ways to choose exactly 1 card: $\binom{10}{1}=10$.
   4. Number of ways to choose exactly 2 cards: $\binom{10}{2}=\dfrac{10\times9}{2}=45$.
   5. Sum of "fewer than 3" selections: $1+10+45=56$.
   6. Selections of at least 3 cards: $1024-56=968$.

   Common Mistakes

   • Forgetting to include the empty selection (size 0) among the cases to subtract.
   • Arithmetic slip computing $\binom{10}{2}$ (it's $45$, not $90$ — remember to divide by $2!$).

   ✓Final answer

   The correct option is (D) — $968$.

   ANSWER: D

3. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.A student has to answer 10 out of 13 questions in an examination choosing atleast 3 from the 5 particular questions. The number of choices available to the student is (A) $196$ (B) $276$ (C) $326$ (D) $156$

   ›Reveal solutionSolution

   Split the 13 questions into the 5 "particular" ones and the other 8, then sum the

   valid combinations where at least 3 of the 10 chosen come from the particular group.

   Answer: (B) 276.

   Concept and Intuition

   "At least 3 from the 5 particular questions" means the number chosen from that group

   of 5 can be $3$, $4$, or $5$ — it cannot be more (only 5 exist) and cannot be less

   (the constraint requires at least 3). For each choice of how many come from the

   particular group, the rest of the 10 must come from the remaining $13-5=8$ ordinary

   questions. Since choosing "which questions" (not order) is what matters, each case is

   a product of two combinations, and the total is the sum over all valid cases (the

   cases are mutually exclusive, so we add rather than multiply across cases).

   Step-by-Step Solution

   1. Let $k$ = number of questions chosen from the 5 particular ones. Need $k\ge 3$ and $k\le 5$ (only 5 available), and the remaining $10-k$ must come from the other 8 questions (which requires $10-k\le 8$, i.e. $k\ge 2$ — automatically satisfied). So $k\in\{3,4,5\}$.
   2. $k=3$: choose 3 of 5 particular and 7 of the other 8: $\binom{5}{3}\binom{8}{7}=10\times 8=80$.
   3. $k=4$: choose 4 of 5 particular and 6 of the other 8: $\binom{5}{4}\binom{8}{6}=5\times 28=140$.
   4. $k=5$: choose all 5 particular and 5 of the other 8: $\binom{5}{5}\binom{8}{5}=1\times 56=56$.
   5. Total $=80+140+56=276$.

   Common Mistakes

   • Forgetting the upper bound $k\le 5$ and trying to include $k=6,7,\dots$, which are impossible since only 5 particular questions exist.
   • Multiplying the three cases together instead of adding them — the cases (choosing exactly 3, exactly 4, exactly 5 from the particular group) are mutually exclusive scenarios, so their counts must be summed.

   ✓Final answer

   The correct option is (B) — $276$.

   ANSWER: B

4. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.The number of ways of distributing 3 dozen fruits (no two fruits are identical) to 9 persons such that each gets the same number of fruits is (A) $\dfrac{36!}{(9!)^4}$ (B) $\dfrac{36!}{(4!)^9}$ (C) ${}^{36}P_9\times 4!$ (D) $\dfrac{36!}{4!(9!)^4}$

   ›Reveal solutionSolution

   This tests the multinomial-coefficient formula for distributing distinct objects into labeled groups of equal size. Answer: $\dfrac{36!}{(4!)^9}$.

   Concept and Intuition

   When distinct objects are split among distinguishable recipients (named people, not anonymous piles), the count is the multinomial coefficient: $\dfrac{n!}{r_1!\,r_2!\cdots r_k!}$, where $r_i$ is how many objects the $i$-th recipient gets. Crucially, since the recipients are distinct people, we do not divide further by $k!$ — that extra division only applies when the groups themselves are unlabeled/interchangeable (e.g., splitting into $k$ identical unlabeled bags).

   Step-by-Step Solution

   1. 3 dozen $=36$ distinct fruits, to be distributed to 9 distinct persons, 4 fruits each ($36/9=4$).
   2. Choose Person 1's 4 fruits from 36: $\binom{36}{4}$ ways.
   3. Choose Person 2's 4 fruits from the remaining 32: $\binom{32}{4}$ ways. Continue this way for all 9 persons.
   4. The product telescopes to $\dfrac{36!}{4!\,32!}\times\dfrac{32!}{4!\,28!}\times\cdots=\dfrac{36!}{(4!)^9}$ (each denominator factorial cancels with the next numerator factorial, leaving only the nine $4!$'s in the denominator).
   5. Since the 9 persons are distinct, this final count is already correct — no further division by $9!$.

   Common Mistakes

   • Dividing by an extra $9!$ (that would be needed only if the 9 groups were unlabeled/identical, which they are not — the persons are distinct).
   • Confusing this with the "identical objects into distinct boxes" stars-and-bars scenario, which doesn't apply here since the fruits are explicitly stated to be distinct.

   ✓Final answer

   The correct option is (B) — $\dfrac{36!}{(4!)^9}$.

   ANSWER: B

5. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.If the number of diagonals of a regular polygon is 35, then the number of sides of the polygon is (A) 12 (B) 9 (C) 10 (D) 11

   ›Reveal solutionSolution

   Apply the standard diagonal-count formula for a polygon and solve the resulting quadratic. Answer: $10$ sides.

   Concept and Intuition

   A polygon with $n$ vertices has $\binom{n}{2}$ total vertex-pairs (lines), of which $n$ are the sides themselves, leaving $\binom{n}{2}-n=\dfrac{n(n-3)}{2}$ diagonals.

   Step-by-Step Solution

   1. Diagonals $=\dfrac{n(n-3)}{2}=35$.
   2. So $n(n-3)=70 \Rightarrow n^2-3n-70=0$.
   3. Factor: we need two numbers multiplying to $-70$ and adding to $-3$: these are $-10$ and $7$. So $(n-10)(n+7)=0$.
   4. $n=10$ or $n=-7$; since $n$ must be a positive integer $\ge3$, $n=10$.

   Common Mistakes

   • Using $\binom{n}{2}$ (total line segments) instead of subtracting the $n$ sides to get diagonals.
   • Accepting the negative root $n=-7$.

   ✓Final answer

   The correct option is (C) — $10$.

   ANSWER: C

6. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.The number of ways of forming the ordered pairs (p, q) such that p > q by choosing p and q from the first 50 natural numbers is (A) 1275 (B) 1250 (C) 1225 (D) 1200

   ›Reveal solutionSolution

   This tests the bijection between unordered pairs and ordered pairs with a strict inequality. Answer: $1225$.

   Concept and Intuition

   Whenever we pick any two distinct numbers $p,q$ from a set, exactly one of $p>q$ or $q>p$ holds — never both, and never neither (since they're distinct). So the number of ordered pairs with $p>q$ is exactly half of all ordered pairs of distinct elements, which is the same as the number of unordered pairs $\binom{n}{2}$.

   Step-by-Step Solution

   1. We need ordered pairs $(p,q)$ with $p,q \in \{1,2,\ldots,50\}$ and $p>q$ (so automatically $p\neq q$).
   2. Choose any 2 distinct numbers from the 50 — this can be done in $\binom{50}{2}$ ways, and for a chosen pair $\{x,y\}$ with $x>y$, there is exactly one ordered pair $(p,q)=(x,y)$ satisfying $p>q$.
   3. $\binom{50}{2} = \dfrac{50\cdot 49}{2} = 25\cdot 49 = 1225$.

   Common Mistakes

   • Confusing this with ordered pairs where repetition or $p\ge q$ is allowed, which would change the count.
   • Trying to count directly via $\sum_{q=1}^{49}(50-q)$ without noticing it's just $\binom{50}{2}$ (though this sum also gives 1225, confirming the result).

   ✓Final answer

   The correct option is (C) — $1225$.

   ANSWER: C

7. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.The number of integers between 10 and 10,000 such that in every integer every digit is greater than its immediate preceeding digit, is (A) 1112 (B) 437 (C) 246 (D) 182

   ›Reveal solutionSolution

   This tests recognizing that "every digit greater than the one before it" forces all digits to be distinct and in increasing order, so each valid number corresponds to exactly one subset of digits — turning a counting-arrangements problem into a simple combinations problem.

   Concept and Intuition

   If every digit must be strictly greater than the digit immediately before it, then all digits in the number must be different (no repeats, since equal digits would violate strict inequality), and once you pick which digits appear, there's only one way to arrange them so they increase — in sorted order. So counting such numbers of a given length $k$ is the same as counting $k$-element subsets of the available digit pool.

   Step-by-Step Solution

   1. Numbers between 10 and 10,000 have 2, 3, or 4 digits.
   2. Digits must be chosen so that each is strictly greater than the previous one. This means all digits are distinct, and their only valid arrangement (increasing) is fixed once the set of digits is chosen.
   3. The leading digit can't be 0 (else it isn't a genuine $k$-digit number), and having 0 anywhere else would violate "digit greater than preceding" unless it's first — but 0 can't be first. So digits must come from $\{1,2,\dots,9\}$.
   4. Count 2-digit numbers: choose 2 digits from 9 → $\binom{9}{2}=36$.
   5. Count 3-digit numbers: choose 3 digits from 9 → $\binom{9}{3}=84$.
   6. Count 4-digit numbers: choose 4 digits from 9 → $\binom{9}{4}=126$.
   7. Total $= 36+84+126 = 246$.

   Common Mistakes

   • Including 0 in the available digit pool, which either breaks the leading-digit rule or the strictly-increasing rule.
   • Trying to count arrangements (permutations) instead of realizing each valid digit-set gives exactly one increasing number.

   ✓Final answer

   The correct option is (C) — 246.

   ANSWER: C

8. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.The number of ways a committee of 8 members can be formed from a group of 10 men and 8 women such that the committee contains at most 5 men and at least 5 women is (A) 8061 (B) 8612 (C) 6082 (D) 8271

   ›Reveal solutionSolution

   Counting committees of 8 from 10 men and 8 women with at most 5 men and at least 5 women gives 8061 by summing over the compatible men-counts.

   Concept and Intuition

   With a fixed committee size, "at least 5 women" is actually the binding constraint here (it forces at most 3 men, which is stricter than "at most 5 men"). Break the count into disjoint cases by the number of men, and use $\binom{10}{\text{men}}\binom{8}{\text{women}}$ for each case, then add.

   Step-by-Step Solution

   1. Committee size is 8, so men + women = 8.
   2. "At least 5 women" $\Rightarrow$ women $\in\{5,6,7,8\}$, i.e. men $\in\{3,2,1,0\}$ — all of which automatically satisfy "at most 5 men", so the real constraint is men $\le 3$.
   3. Case men=0, women=8: $\binom{10}{0}\binom{8}{8}=1\times1=1$.
   4. Case men=1, women=7: $\binom{10}{1}\binom{8}{7}=10\times8=80$.
   5. Case men=2, women=6: $\binom{10}{2}\binom{8}{6}=45\times28=1260$.
   6. Case men=3, women=5: $\binom{10}{3}\binom{8}{5}=120\times56=6720$.
   7. Total $=1+80+1260+6720=8061$.

   Common Mistakes

   • Treating "at most 5 men" as the binding constraint and missing that "at least 5 women" is actually tighter.
   • Arithmetic slips in the binomial coefficients or the final sum.

   ✓Final answer

   The correct option is (A) — 8061.

   ANSWER: A

9. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
   Q.If a polygon of $n$ sides has $275$ diagonals, then $n$ is (A) $25$ (B) $35$ (C) $20$ (D) $15$

   ›Reveal solutionSolution

   Using the standard diagonal-count formula $\dfrac{n(n-3)}{2}=275$ and solving the resulting quadratic gives $n=25$.

   Concept and Intuition

   A convex polygon with $n$ vertices has $\binom{n}{2}$ total vertex-pairs, of which $n$ are sides (adjacent pairs) — so the number of diagonals is $\binom{n}{2}-n = \dfrac{n(n-1)}{2}-n=\dfrac{n(n-3)}{2}$.

   Step-by-Step Solution

   1. Diagonals formula: $\dfrac{n(n-3)}{2}=275$.
   2. Multiply both sides by 2: $n(n-3)=550 \Rightarrow n^2-3n-550=0$.
   3. Apply the quadratic formula: $n = \dfrac{3\pm\sqrt{9+4(550)}}{2} = \dfrac{3\pm\sqrt{2209}}{2}$.
   4. $\sqrt{2209}=47$ (since $47^2=2209$), so $n=\dfrac{3+47}{2}=25$ (rejecting the negative root, since $n$ must be a positive integer $\ge3$).
   5. Check: $\dfrac{25\cdot22}{2}=\dfrac{550}{2}=275$ ✓.

   Common Mistakes

   • Using the wrong diagonal formula (e.g. confusing it with $\binom{n}{2}$, the total number of line segments including sides).
   • Forgetting to reject the negative root of the quadratic.

   ✓Final answer

   The correct option is (A) — $25$.

   ANSWER: A

10. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
    Q.The number of ways in which 17 apples can be distributed among four guests such that each guest gets at least 3 apples is (A) $1140$ (B) $336$ (C) $36$ (D) $56$

    ›Reveal solutionSolution

    Guarantee the minimum first, then distribute the remainder with stars-and-bars. Answer: $56$.

    Concept and Intuition

    When every recipient needs at least a fixed minimum, the standard trick is to hand out that minimum up front, reducing the problem to an ordinary non-negative-integer distribution, solvable by stars-and-bars: the number of ways to write $n$ as an ordered sum of $k$ non-negative integers is $\binom{n+k-1}{k-1}$.

    Step-by-Step Solution

    1. Let $x_1,x_2,x_3,x_4\ge3$ be the apples each guest gets, with $x_1+x_2+x_3+x_4=17$.
    2. Substitute $y_i=x_i-3\ge0$: then $y_1+y_2+y_3+y_4=17-12=5$.
    3. Number of non-negative integer solutions: $\binom{5+4-1}{4-1}=\binom83=56$.

    Common Mistakes

    • Forgetting to subtract the guaranteed minimum before applying stars-and-bars.
    • Using $\binom{n+k-1}{k}$ instead of $\binom{n+k-1}{k-1}$ (same value here since $\binom83=\binom85$, but easy to mix up the reasoning).

    ✓Final answer

    The correct option is (D) — $56$.

    ANSWER: D

11. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
    Q.There were two women participating with some men in a chess tournament. Each participant played two games with the other. The number of games that the men played between themselves is 66 more than that of the men played with the women. Then the total number of participants in the tournament is (A) 17 (B) 13 (C) 11 (D) 19

    ›Reveal solutionSolution

    This is a counting problem where each pair of participants plays two games; setting up the men-men vs men-women game counts gives $n=11$ men, so $13$ participants total.

    Concept and Intuition

    With $n$ men and 2 women, and every pair of participants playing 2 games between them, the number of games played between any two groups is $2\times(\text{number of pairs across the groups})$. Setting up "games among men" minus "games between men and women" as the given excess of 66 turns the word problem into a quadratic in $n$.

    Step-by-Step Solution

    1. Games played among the $n$ men alone: each of the $\binom{n}{2}$ pairs plays 2 games, so total $=2\binom{n}{2}=n(n-1)$.
    2. Games played between men and the 2 women: each of the $n$ men plays each of the 2 women twice, so total $=n\times2\times2=4n$.
    3. Given: $n(n-1) = 4n+66$.
    4. $n^2-n-4n-66=0 \Rightarrow n^2-5n-66=0$.
    5. Solve: $n=\dfrac{5\pm\sqrt{25+264}}{2}=\dfrac{5\pm17}{2}$, taking the positive root $n=11$.
    6. Total participants $= 11\text{ men} + 2\text{ women} = 13$.

    Common Mistakes

    • Forgetting the factor of 2 for "each pair plays two games," undercounting both game totals by half.
    • Miscounting men-women games as $2n$ instead of $4n$ (each of $n$ men plays each of 2 women, twice each).

    ✓Final answer

    The correct option is (B) — 13.

    ANSWER: B

12. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
    Q.The number of non-negative integral solutions of $x_1 + x_2 + x_3 + x_4 = 10$ is (A) $120$ (B) $144$ (C) $256$ (D) $286$

    ›Reveal solutionSolution

    This is the classic stars-and-bars count: $\binom{n+r-1}{r-1}$ solutions for $r$ non-negative variables summing to $n$. Here $\binom{13}{3}=286$. Answer: (D).

    Concept and Intuition

    Distributing $n$ identical units among $r$ variables (allowing zero) is equivalent to arranging $n$ stars and $r-1$ bars in a row.

    Step-by-Step Solution

    1. $n=10$, $r=4$.
    2. Count $= \binom{n+r-1}{r-1} = \binom{13}{3}$.
    3. $\binom{13}{3} = \dfrac{13\times12\times11}{6} = \dfrac{1716}{6}=286$.

    Common Mistakes

    • Confusing this with the positive-integer-solutions formula $\binom{n-1}{r-1}$, wrongly giving $84$.
    • Misidentifying $n$ or $r$.

    ✓Final answer

    The correct option is (D) — $286$.

    ANSWER: D