Q.The number of possible outcomes when a coin is tossed $6$ times is
(A) $36$
(B) $64$
(C) $12$
(D) $32$

Showing the 12 most recent of 16 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.The number of times the digit 5 occurs in all the numbers from 1 to 1000 is (A) $243$ (B) $297$ (C) $300$ (D) $305$

   ›Reveal solutionSolution

   Padding all numbers to 3 digits shows each digit position cycles through 0–9 equally often (100 times each), so digit 5 appears $300$ times total from 1 to 1000.

   Concept and Intuition

   Counting digit occurrences across a full block like 1–999 is easiest by imagining every number written with leading zeros as a 3-digit string (000 to 999). In such a complete, uniform block, each of the three positions independently takes every value 0–9 exactly $1000/10=100$ times, since the block is perfectly symmetric in each digit slot.

   Step-by-Step Solution

   1. Write all integers from 1 to 999 as 3-digit strings with leading zeros (000 to 999); this doesn't change how many actual 5's appear (leading zeros are never 5's).
   2. In the units place, as the string runs from 000 to 999, each digit 0–9 appears exactly 100 times, so digit 5 appears 100 times in the units position.
   3. By the same symmetry, digit 5 appears exactly 100 times in the tens position and 100 times in the hundreds position. …

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2. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.There are 15 identical balls for sale among which 4 are red, 5 are black and 6 are white. If a person buys at least one ball out of these 15 balls, then the total number of ways in which that person can buy the balls is (A) $120$ (B) $119$ (C) $210$ (D) $209$

   ›Reveal solutionSolution

   With identical balls of each colour, counting "how many of each colour to take" (0 up to the max) gives $(n_1+1)(n_2+1)(n_3+1)$ total combinations including buying nothing; subtracting the "buy nothing" case gives the answer.

   Concept and Intuition

   When items within a colour are identical, a selection is fully described just by how many of each colour you take — there's no further choice of "which" ball, since they're indistinguishable. For a colour with $n$ identical balls, you can take $0, 1, 2, \ldots, n$ of them: that's $n+1$ possibilities. Since choices for different colours are independent, multiply the possibilities together, then remove the single case where you take zero of every colour (since the person must buy at least one ball overall).

   Step-by-Step Solution

   1. Red balls: 4 identical, so $0$ to $4$ can be taken → $5$ choices.
   2. Black balls: 5 identical, so $0$ to $5$ can be taken → $6$ choices.
   3. White balls: 6 identical, so $0$ to $6$ can be taken → $7$ choices. …

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3. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.The number of all possible positive integral solutions of the equation $xyz = 30$ is (A) $24$ (B) $25$ (C) $26$ (D) $27$

   ›Reveal solutionSolution

   Since $30=2\times3\times5$ has three distinct prime factors each to the first power, each prime independently 'chooses' one of $x,y,z$ to belong to, giving $3^3=27$ ordered positive-integer solutions.

   Concept and Intuition

   When counting ordered triples $(x,y,z)$ of positive integers with a given product $N$, the key idea is to distribute each prime power of $N$'s factorization among the three variables independently. If a prime appears with exponent $e$ in $N$, the number of ways to split that exponent among 3 variables is $\binom{e+2}{2}$ (stars and bars) — but when $e=1$ (as here, for every prime in $30$), that formula gives simply $3$ (the whole prime goes to one of the three slots, no splitting possible).

   Step-by-Step Solution

   1. Factorize $30=2^1\times3^1\times5^1$ — three distinct primes, each with exponent exactly $1$.
   2. For the prime $2$: since its exponent is $1$, it cannot be split between two of $x,y,z$ — it must belong entirely to exactly one of them. There are $3$ choices (it goes to $x$, or to $y$, or to $z$).
   3. Same reasoning applies independently to the prime $3$: $3$ choices.
   4. And independently to the prime $5$: $3$ choices.
   5. Since the choices for the three primes are independent of each other, the total number of ordered triples is $3\times3\times3=3^3=27$. …

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4. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.An unbiased coin is tossed 8 times. The probability that head appears consecutively at least 5 times is (A) $\dfrac{5}{256}$ (B) $\dfrac{5}{128}$ (C) $\dfrac{5}{64}$ (D) $\dfrac{5}{32}$

   ›Reveal solutionSolution

   Exactly $20$ of the $256$ sequences contain a run of $\ge 5$ heads, so the probability is $20/256=\dfrac{5}{64}$.

   Concept and Intuition

   Each toss is independent, giving $2^8=256$ equally likely head/tail strings. "Head appears consecutively at least 5 times" means the string contains a block of $5$ or more successive $H$s. We count such strings by inclusion-exclusion on where the $5$-block starts.

   Step-by-Step Solution

   1. Let $A_i$ be the event that positions $i,\dots,i+4$ are all heads, $i=1,2,3,4$. Each fixes $5$ positions leaving $3$ free: $|A_i|=2^3=8$, so $\sum|A_i|=32$.
   2. Pairwise overlaps (union of two blocks) contribute $4+2+1+4+2+4=17$.
   3. Triple overlaps contribute $2+1+1+2=6$; the quadruple overlap contributes $1$.
   4. Inclusion-exclusion: favourable $=32-17+6-1=20$.
   5. Check by the complement: strings of length 8 with no run of $5$ heads number $236$, and $256-236=20$ agrees.
   6. Probability $=\dfrac{20}{256}=\dfrac{5}{64}$.

   Common Mistakes …

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5. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
   Q.A string of letters is to be formed by using 4 letters from all the letters of the word "MATHEMATICS". The number of ways this can be done such that two letters are of same kind and the other two are of different kind is (A) $756$ (B) $252$ (C) $840$ (D) $360$

   ›Reveal solutionSolution

   Count 4-letter strings from MATHEMATICS with exactly one repeated pair and two other distinct letters; total is $756$.

   Concept and Intuition

   MATHEMATICS has 11 letters: M,A,T,H,E,M,A,T,I,C,S. Distinct letters are M,A,T,H,E,I,C,S (8 total), where M, A, T each occur twice and H,E,I,C,S occur once. "Two of the same kind, two of different kind" means we pick one letter to repeat (must come from the 2-count letters) and two more, mutually distinct, single-occurrence-in-the-string letters.

   Step-by-Step Solution

   1. Choose which letter repeats: only M, A, or T can repeat (each has 2 copies available) — $3$ choices.
   2. Choose the other two distinct letters from the remaining $7$ distinct letters (8 total minus the one already chosen to repeat): $\binom{7}{2}=21$ ways. …

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6. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
   Q.If set A contains 8 elements then number of subsets of A which contain at least 6 elements is (A) 28 (B) 73 (C) 37 (D) 82

   ›Reveal solutionSolution

   This tests counting subsets by size using combinations, then summing over all sizes $\ge 6$. The answer is 37.

   Concept and Intuition

   The number of $k$-element subsets of an $n$-element set is $\binom{n}{k}$. "At least 6 elements" is not a single size — it is the union of the size-6, size-7, and size-8 cases, so we add the three combination counts.

   Step-by-Step Solution

   1. Here $n=8$. We need subsets of size $6$, $7$, or $8$.
   2. $\binom{8}{6} = \binom{8}{2} = \dfrac{8\times7}{2} = 28$ (using $\binom{n}{k}=\binom{n}{n-k}$).
   3. $\binom{8}{7} = \binom{8}{1} = 8$.
   4. $\binom{8}{8} = 1$ (the full set itself). …

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7. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
   Q.A test containing 3 objective type of questions is conducted in a class. Each question has 4 options and only one option is the correct answer. No two students of the class have answered identically and no student has written all correct answers. If every student has attempted all the questions, then the maximum possible number of students who has written the test is (A) $80$ (B) $63$ (C) $15$ (D) $11$

   ›Reveal solutionSolution

   Count all possible answer-patterns, remove the one pattern that's banned (all correct), and that's the maximum number of distinct students. Answer: $63$.

   Concept and Intuition

   Each student's set of answers to the 3 questions is a string from an alphabet of 4 options per question — a simple counting problem, not really about correctness at all except for one excluded case.

   Step-by-Step Solution

   1. Total possible answer-patterns across 3 questions, 4 options each: $4\times4\times4=64$.
   2. "No student has written all correct answers" removes exactly 1 pattern (the fully-correct one): $64-1=63$ remain usable.
   3. "No two students answered identically" means each usable pattern can be used by at most one student.
   4. So the maximum number of students is $63$ (one student per remaining pattern). …

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8. AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
   Q.Among the 4 digit numbers formed using the digits 0, 1, 2, 3, 4 when repetition of digits allowed, the number of numbers which are divisible by 4 is (A) $140$ (B) $160$ (C) $180$ (D) $200$

   ›Reveal solutionSolution

   Divisibility by 4 depends only on the last two digits; count valid last-two-digit pairs, then multiply by the free choices for the first two digits.

   Concept and Intuition

   Since $100$ is divisible by $4$, any number $N = 100\cdot(\text{leading part}) + (\text{last two digits})$ is divisible by $4$ exactly when its last two digits (as a two-digit number, possibly with a leading zero, e.g. $04$) are divisible by $4$. This turns a 4-digit divisibility question into a small, checkable 2-digit sub-problem.

   Step-by-Step Solution

   1. The 4-digit number is $d_1d_2d_3d_4$ with each $d_i \in \{0,1,2,3,4\}$ (repetition allowed) and $d_1 \ne 0$.
   2. Divisibility by 4 needs $10d_3 + d_4 \equiv 0 \pmod 4$, i.e. $2d_3 + d_4 \equiv 0 \pmod 4$ (since $10 \equiv 2 \pmod 4$).
   3. Check each $d_3 \in \{0,1,2,3,4\}$:
      • $d_3=0$: need $d_4\equiv0\ (\text{mod }4)$: $d_4 \in\{0,4\}$ — 2 options.
      • $d_3=1$: need $d_4\equiv2$: $d_4=2$ — 1 option.
      • $d_3=2$: need $d_4\equiv0$: $d_4\in\{0,4\}$ — 2 options.
      • $d_3=3$: need $d_4\equiv2$: $d_4=2$ — 1 option. …

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9. AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
   Q.If a seven digit number formed with distinct digits $4, 6, 9, 5, 3, x$ and $y$ is divisible by $3$, then the number of such ordered pairs $(x, y)$ is (A) $7$ (B) $8$ (C) $9$ (D) $10$

   ›Reveal solutionSolution

   Since the five fixed digits already sum to a multiple of $3$, we just need $x+y\equiv0\pmod3$ from the remaining pool $\{0,1,2,7,8\}$ — sorting by residue class gives exactly $8$ ordered pairs.

   Concept and Intuition

   A number is divisible by $3$ iff its digit sum is divisible by $3$. With five digits fixed, only the residue of $x+y$ modulo $3$ matters — so classify the remaining candidate digits by their residue mod $3$ and count valid combinations.

   Step-by-Step Solution

   1. Sum of fixed digits: $4+6+9+5+3=27$, and $27\equiv0\pmod3$.
   2. So we need $x+y\equiv0\pmod3$ as well (so the total stays $\equiv0$).
   3. Digits $0$–$9$ excluding the five already used ($3,4,5,6,9$) leave $\{0,1,2,7,8\}$ as the pool for $x,y$ (must be distinct from each other and from the fixed digits).
   4. Residues mod 3: $0\to0$; $1\to1$; $2\to2$; $7=6+1\to1$; $8=6+2\to2$.
   5. So residue classes are: $\{0\}$ (residue 0, just one element), $\{1,7\}$ (residue 1), $\{2,8\}$ (residue 2).
   6. Pairs summing to $0\bmod3$: need residue $0+0$ (impossible — only one element available in that class, can't pick two distinct), or residue $1+2$ (valid). …

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10. AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
    Q.The number of odd positive divisors of 67500 is (A) $16$ (B) $18$ (C) $20$ (D) $22$

    ›Reveal solutionSolution

    Factorising 67500 = 2²·3³·5⁴ and counting divisors of the odd part (3³·5⁴) gives 4×5=20 odd divisors.

    Concept and Intuition

    To count only the odd divisors of a number, strip out all factors of 2 from its prime factorisation, then count divisors of what remains using the standard (exponent+1) product rule.

    Step-by-Step Solution

    1. Factorise 67500: 67500 = 675 × 100 = (27×25) × (4×25) = 3³×5² × 2²×5² = 2² × 3³ × 5⁴.
    2. Verify: 2²=4, 3³=27, 5⁴=625; 4×27=108; 108×625=67500 ✓.
    3. Odd divisors come only from the odd part 3³×5⁴ (ignore the 2² factor entirely).
    4. Number of divisors of 3³×5⁴ = (3+1)(4+1) = 4×5 = 20. …

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11. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
    Q.Number of four digit numbers that can be formed using all the digits except zero such that every number has exactly 2 distinct digits in it is (A) 189 (B) 216 (C) 288 (D) 504

    ›Reveal solutionSolution

    Count the digit-pairs, then count the strings that use both digits of a pair at least once — the answer is (D) 504.

    Concept and Intuition

    "Exactly 2 distinct digits" in a 4-digit number means: pick which 2 digits are used, then fill 4 positions with only those two digits such that both actually appear (not all four positions the same digit).

    Step-by-Step Solution

    1. Since 0 is excluded, digits come from $\{1,2,\dots,9\}$ (9 digits).
    2. Choose the 2 digits to be used: $^9C_2 = 36$.
    3. Each of the 4 positions can independently be either of the 2 chosen digits: $2^4=16$ total strings.
    4. Remove the 2 "monotone" strings where all 4 digits are the same (all digit-1 or all digit-2): $16-2=14$. …

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12. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
    Q.The number of ways of distributing 500 dissimilar boxes equally among '50' persons is (A) $500!/(10!)^{50}.50!$ (B) $500!/(50!)^{10}.10!$ (C) $500!/(50!)^{10}$ (D) $500!/(10!)^{50}$

    ›Reveal solutionSolution

    Distributing distinct objects into labeled groups of equal size is a multinomial coefficient — the answer is (D) $500!/(10!)^{50}$.

    Concept and Intuition

    When items are distinct AND the recipients (groups) are distinct/labeled (here, 50 named persons), the count is the multinomial coefficient $\dfrac{n!}{(n/k)!^k}$ with no extra division by $k!$. Dividing further by $k!$ is only needed when the groups themselves are indistinguishable (e.g., "split into 50 unlabeled piles").

    Step-by-Step Solution

    1. 500 distinct boxes are to be split into 50 groups of 10 each (since $500/50=10$), one group per named person.
    2. The number of ways to partition $n$ distinct items into labeled groups of sizes $n_1,\dots,n_k$ is $\dfrac{n!}{n_1!\,n_2!\cdots n_k!}$. …

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