Q.If the set $A$ contains 5 elements and the set $B$ contains 6 elements, then the number of one-one and onto mappings from $A$ to $B$ is
(A) 720
(B) 120
(C) 0
(D) none of these

Showing the 12 most recent of 17 on this concept.

1. AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
   Q.If a set $A$ has m elements and set $B$ has $n$ elements and the number of injections from $A$ to $B$ is $2520$. Then $m$ is equal to (A) $2$ (B) $7$ (C) $6$ (D) $5$

   ›Reveal solutionSolution

   $2520=7\times6\times5\times4\times3={}^7P_5$, so the number of elements in $A$ is $m=5$.

   Concept and Intuition

   An injection (one-one function) from a set of size $m$ into a set of size $n$ picks an ordered sequence of $m$ distinct elements from the $n$ available — exactly $^nP_m=n(n-1)(n-2)\cdots(n-m+1)$, a product of $m$ consecutive descending integers.

   Step-by-Step Solution

   1. We need $^nP_m = n!/(n-m)! = 2520$ for some positive integers $m<n$ (as one of the answer options).
   2. Factor $2520 = 7\times6\times5\times4\times3$ — a product of exactly $5$ consecutive integers starting from $7$ downward.
   3. This matches $^7P_5 = 7\times6\times5\times4\times3=2520$, i.e., $n=7$, $m=5$. …

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2. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
   Q.If set A has 5 elements, set B has 7 elements then the number of many one functions that can be defined from A to B is (A) $7^5 - 7$ (B) $5^7 - 5$ (C) $5^7 - {}^7P_5$ (D) $7^5 - {}^7P_5$

   ›Reveal solutionSolution

   Many-one functions are simply all functions minus the injective (one-one) ones. Answer: $7^5-{}^7P_5$.

   Concept and Intuition

   Every function from a 5-element set to a 7-element set is either one-one or many-one (some element of $B$ has more than one pre-image, or equivalently the function is not injective). So many-one count = total functions − injective functions.

   Step-by-Step Solution

   1. Total number of functions from $A$ (5 elements) to $B$ (7 elements) $=7^5$ (each of the 5 elements of $A$ can map to any of 7 elements of $B$).
   2. Since $|A|<|B|$, one-one functions exist and their count is ${}^7P_5 = \dfrac{7!}{2!}$. …

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3. AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
   Q.If a set A has m-elements and the set B has n-elements, then the number of injections from A to B is (A) ${}^{n}C_{m}$ if $n \geq m$ (B) ${}^{n}P_{m}$ if $n \geq m$ (C) $0$ if $n \geq m$ (D) $m \cdot {}^{n}C_{m}$ if $n \geq m$

   ›Reveal solutionSolution

   Counting injections is equivalent to choosing and ordering $m$ distinct images out of $n$, giving $^{n}P_{m}$ — but only possible at all when $n\ge m$.

   Concept and Intuition

   An injective (one-to-one) function from $A$ (with $m$ elements) to $B$ (with $n$ elements) must map every element of $A$ to a distinct element of $B$. This is exactly the same combinatorial structure as arranging $m$ distinct objects into $n$ distinct positions without repetition — a permutation problem, not a combination problem (since which element of $A$ maps to which specific element of $B$ matters, unlike choosing a subset).

   Step-by-Step Solution

   1. By the pigeonhole principle, if $n<m$, no injective function can exist (there aren't enough distinct targets for all $m$ elements), so the count is $0$ in that case.
   2. When $n\ge m$, build an injection element-by-element: the first element of $A$ can map to any of $n$ elements of $B$; the second element of $A$ can map to any of the remaining $n-1$ elements (to keep injectivity); and so on, down to the $m^{th}$ element having $n-m+1$ choices. …

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4. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.Let p denote the number of surjections from a set containing 6 elements to a set containing 2 elements. Let q denote the number of injections from a set containing 3 elements to a set containing 5 elements. Let r denote the number of bijections from a set containing 4 elements to itself. Then $p - q + r =$ (A) $22$ (B) $98$ (C) $26$ (D) $146$

   ›Reveal solutionSolution

   The problem asks for $p - q + r$, where $p$ counts surjections from a 6‑element set to a 2‑element set, $q$ counts injections from a 3‑element set to a 5‑element set, and $r$ counts bijections from a 4‑element set to itself. We compute each separately: $p = 2^6 - 2 = 62$, $q = 5 \times 4 \times 3 = 60$, $r = 4! = 24$. Then $p - q + r = 62 - 60 + 24 = 26$. The correct option is (C).

   ---

   Concept and Intuition: Counting Functions

   We are counting three types of functions between finite sets. The key idea is that each type — surjection, injection, bijection — imposes a different condition on how elements of the domain map to the codomain.

   • Surjection: every element of the codomain must be hit at least once. For a small codomain (size 2), we can count total functions and subtract those that miss at least one element.
   • Injection: no two domain elements map to the same codomain element. This is like choosing an ordered list of distinct images.
   • Bijection: both surjective and injective; for a set to itself, this is just a permutation.

   We compute each separately, then combine.

   ---

   Step‑by‑Step Solution

   1. Compute $p$: number of surjections from a 6‑element set to a 2‑element set.

   Let the domain have 6 elements and the codomain have 2 elements, say $\{a, b\}$.

   A function is surjective if both $a$ and $b$ are images of at least one domain element.

   • Total number of functions from a 6‑element set to a 2‑element set: each of the 6 elements has 2 choices, so $2^6 = 64$.
   • Subtract functions that are not surjective. A function fails to be surjective if it misses at least one codomain element.
     • Miss $a$: all 6 elements map to $b$ → 1 function.
     • Miss $b$: all 6 elements map to $a$ → 1 function.
     • These two cases are disjoint (cannot miss both because the codomain has only 2 elements).
   • So number of non‑surjective functions = $1 + 1 = 2$.

   Thus

   $p = 2^6 - 2 = 64 - 2 = 62.$

   Tip

   For a codomain of size 2, the formula $2^n - 2$ counts surjections from an $n$-element set. This works because the only way to fail surjectivity is to map everything to a single element.

   --- …

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5. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.Let $A$ and $B$ be two non-empty sets with $n(A) = 4$ and $n(B) = 5$. If a mapping is selected at random from the set of all mapping from $A$ to $B$, then the probability of getting a many-one mapping is (A) $\dfrac{101}{125}$ (B) $\dfrac{96}{125}$ (C) $\dfrac{921}{1024}$ (D) $\dfrac{113}{128}$

   ›Reveal solutionSolution

   Many-one is the complement of one-one among all functions $A\to B$. Answer: $\dfrac{101}{125}$.

   Concept and Intuition

   Every function from $A$ to $B$ is either one-one (injective) or many-one (not injective) — these are complementary events, so $P(\text{many-one}) = 1 - P(\text{one-one})$.

   Step-by-Step Solution

   1. Total number of functions from a 4-element set to a 5-element set: each of the 4 elements of $A$ can map to any of 5 elements of $B$, giving $5^4 = 625$.
   2. Number of one-one (injective) functions: choose images one at a time without repetition: $5\times4\times3\times2 = 120$.
   3. $P(\text{one-one}) = \dfrac{120}{625} = \dfrac{24}{125}$. …

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6. AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
   Q.If a set A has n elements, then the number of functions defined from A to A that are not one-one is (A) $(n)^{n^2}$ (B) $n! - (^{n}C_0 + {}^{n}C_1 + {}^{n}C_2 + \ldots + {}^{n}C_n)$ (C) $n^n - n!$ (D) $n^n$

   ›Reveal solutionSolution

   Total functions from A to A number n^n; the one-one ones (which, on a finite set to itself, are exactly the bijections) number n!; subtracting gives n^n - n! not-one-one functions.

   Concept and Intuition

   A function $f: A \to A$ assigns each of the n elements of A to one of n possible images, independently, giving $n^n$ total functions. On a finite set mapped to itself, injective (one-one) iff surjective (onto) - a standard pigeonhole fact. So one-one functions are precisely the bijections, and there are n! of these.

   Step-by-Step Solution

   1. Total functions $A \to A$: each domain element has n choices, total $= n^n$.
   2. On a finite set A to itself, one-one $\iff$ bijection. …

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7. AP EAPCET 2021Set eng-2021-08-25-FN1 markMCQ
   Q.Let $A = \{1, 2, 3, 4, 5, 6\}$. Number of functions '$f$' from $A$ to $A$ such that $f(m) + f(n) = 7$ whenever $m + n = 7$ is ______ (A) 525 (B) 216 (C) 200 (D) 729

   ›Reveal solutionSolution

   The constraint splits $A$ into 3 independent complementary pairs; each pair allows 6 free choices, giving $6^3=216$ valid functions.

   Concept and Intuition

   The condition "$f(m)+f(n)=7$ whenever $m+n=7$" only links elements that sum to 7. Listing all such pairs in $\{1,...,6\}$ shows they partition the set completely into three disjoint pairs, with no element paired to itself (since $7$ is odd, no $m=n$ solves $2m=7$). Within each pair, fixing the value of $f$ on one element automatically determines $f$ on its partner (as $7$ minus that value), and this determined value is guaranteed to also lie in $\{1,...,6\}$. So the three pairs act as three completely independent "slots," each with 6 free choices.

   Step-by-Step Solution

   1. List all pairs $(m,n)$ with $m+n=7$, $m,n\in\{1,...,6\}$: $(1,6),(2,5),(3,4)$ — exactly 3 disjoint pairs covering all 6 elements.
   2. For pair $\{1,6\}$: choose $f(1)\in\{1,...,6\}$ freely (6 choices); then $f(6)=7-f(1)$ is forced (and automatically valid, since $7-f(1)\in\{1,...,6\}$ whenever $f(1)$ is). …

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8. AP EAPCET 2022Set eng-2022-07-06-AN1 markMCQ
   Q.How many numbers between 10 and 10,000 can be formed by using the digits 1,2,3,4,5, if no digit is repeated in any number? (A) 200 (B) 775 (C) 60 (D) 120

   ›Reveal solutionSolution

   This tests counting numbers with $k$ digits (for $k=2,3,4$) formed without repetition from a digit set that has no zero, then summing over the valid digit-lengths. Answer: 200.

   Concept and Intuition

   Since the numbers must lie strictly between 10 and 10,000, they can have 2, 3, or 4 digits (a 1-digit number can't exceed 10, and a 5-digit number using all 5 given digits would be at least 12345, far above 10,000... actually a 5-digit number itself exceeds 10,000, so only 2,3,4-digit numbers qualify). Because the available digits are 1 through 5 (no zero), every arrangement automatically forms a valid number with no leading-zero issue, so this is a straightforward permutation count for each length, summed together.

   Step-by-Step Solution

   1. Numbers strictly between 10 and 10,000 can have 2, 3, or 4 digits (1-digit numbers are $\le 9<10$; 5-digit numbers using digits 1-5 are $\ge 12345 > 10{,}000$).
   2. Digits available: $\{1,2,3,4,5\}$, each used at most once per number (no repetition), and none is zero, so leading-zero is never an issue.
   3. Count of 2-digit numbers: choose and arrange 2 digits from 5: $P(5,2)=5\times4=20$.
   4. Count of 3-digit numbers: $P(5,3)=5\times4\times3=60$.
   5. Count of 4-digit numbers: $P(5,4)=5\times4\times3\times2=120$.
   6. Total $=20+60+120=200$.

   Common Mistakes …

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9. AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQ
   Q.In how many ways can 5 balls be placed in 4 tins if any number of balls can be placed in any tin? (A) ${}^5P_4$ (B) ${}^5C_4$ (C) $4^5$ (D) $5^4$

   ›Reveal solutionSolution

   With distinct balls and unrestricted tin capacity, each ball independently has 4 choices, giving $4^5$ total ways.

   Concept and Intuition

   This is the classic "functions from a set to a set" count: placing $n$ distinguishable balls into $k$ distinguishable tins with no restriction on how many balls a tin can hold is equivalent to choosing, for each ball, which tin it goes into — independently. That is $k^n$, not a permutation or combination formula (those would apply if each tin could hold at most one ball).

   Step-by-Step Solution

   1. Ball 1 can go into any of the 4 tins: 4 choices.
   2. Ball 2 can independently go into any of the 4 tins: 4 choices.
   3. This repeats independently for all 5 balls (since any number of balls may occupy a tin). …

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10. AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
    Q.The number of all four digit numbers which begin with 4 and end with either zero or five is (A) 200 (B) 64 (C) 256 (D) 32

    ›Reveal solutionSolution

    Fix the first digit as $4$ and the last as $0$ or $5$; the two middle digits are unrestricted, giving $1\times10\times10\times2=200$ numbers.

    Concept and Intuition

    A straightforward application of the multiplication principle for counting arrangements with some positions fixed/restricted and others free.

    Step-by-Step Solution

    1. Four-digit number: positions are (thousands, hundreds, tens, units).
    2. Thousands digit is fixed at $4$: $1$ way.
    3. Units digit must be $0$ or $5$: $2$ ways.
    4. Hundreds and tens digits are each any of $0$–$9$: $10$ ways each. …

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11. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
    Q.The number of numbers lying between 1000 and 10000 such that every number contains the digits 3 and 7 only once without repetition is (A) $1140$ (B) $918$ (C) $720$ (D) $810$

    ›Reveal solutionSolution

    Split into "other two digits equal" and "other two digits different", handle the leading-zero restriction in each, and add. Answer: $720$.

    Concept and Intuition

    "Contains 3 and 7 only once, without repetition" restricts repetition of the digits $3$ and $7$ specifically, but doesn't forbid the other two digits (which must avoid $3,7$) from matching each other. So we must count both the case where those two extra digits are the same and the case where they differ, and in each subtract numbers that would start with $0$.

    Step-by-Step Solution

    1. Other two digits equal (multiset $\{3,7,e,e\}$, $e\in\{0,1,2,4,5,6,8,9\}$): number of distinct arrangements of each multiset is $\dfrac{4!}{2!}=12$. Over 8 choices of $e$: $8\times12=96$.
    2. Subtract leading-zero cases: only happens when $e=0$ (multiset $\{3,7,0,0\}$). Fixing one $0$ in front and arranging the remaining $\{3,7,0\}$ (all distinct) in the last 3 places gives $3!=6$ invalid numbers. So valid count here $=96-6=90$.
    3. Other two digits distinct ($d_1\ne d_2$, both from the 8-digit set): choose the pair, $\binom82=28$ ways; arrange all 4 distinct digits $\{3,7,d_1,d_2\}$ in $4!=24$ ways: raw total $28\times24=672$. …

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12. AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
    Q.The number of all 8 digit odd numbers is (A) $45 \times 10^6$ (B) $90 \times 10^6$ (C) $9 \times 10^8$ (D) $9 \times 10^6$

    ›Reveal solutionSolution

    Counting 8-digit odd numbers by the multiplication principle: first digit $9$ options, last digit $5$ options (odd), middle six digits $10$ each, giving $45\times10^6$.

    Concept and Intuition

    This is a straightforward fundamental-counting-principle problem: fix the constrained positions (first digit can't be $0$; last digit must be odd) and multiply by the free choices for the rest.

    Step-by-Step Solution

    1. First digit (leftmost, to make it an 8-digit number): can be $1$ through $9$ → $9$ choices.
    2. Last digit (to make the number odd): must be one of $1,3,5,7,9$ → $5$ choices. …

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