Q.In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?

Showing the 12 most recent of 53 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.The number of ways of arranging 10 men and 5 women around a circular table such that no two women sit together is (A) $10!$ (B) $5!$ (C) $10! \, 5!$ (D) $9! \, {}^{10}P_5$

   ›Reveal solutionSolution

   This tests circular arrangement with a non-adjacency restriction, solved by the standard "fix the majority, use their gaps" method. Answer: $9!\,{}^{10}P_5$.

   Concept and Intuition

   Whenever a problem asks for a circular arrangement where certain people (here, the 5 women) must not sit next to each other, the standard technique is: arrange everyone else around the circle first, which automatically creates gaps between them, and then insert the restricted people into distinct gaps so they can never be adjacent to each other.

   Step-by-Step Solution

   1. Arrange the 10 men around the circular table. A circular arrangement of $n$ distinct people has $(n-1)!$ ways (since rotations of the same arrangement are identical), so this gives $9!$ ways.
   2. Once the 10 men are seated in a circle, there are exactly 10 gaps between consecutive men (one gap after each man, going around the table). …

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2. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.If all possible 6 digit numbers are formed by using all the digits $1, 3, 5, 6, 7, 9$ without repeating any digit, then the number of numbers which are greater than $3,00,000$ and divisible by $4$ is (A) $5 \times 4!$ (B) $5! \times 4!$ (C) $13 \times 3!$ (D) $4 \times 4!$

   ›Reveal solutionSolution

   Divisibility by 4 forces the units digit to be $6$ here; then excluding leading digit $1$ (to keep the number above $300000$) leaves $4\times4!=96$ valid numbers.

   Concept and Intuition

   A number is divisible by 4 exactly when its last two digits form a number divisible by 4. With only six specific digits available and no repetition, we first pin down which digit pairs at the end can give a multiple of 4, then count arrangements of the rest subject to the "greater than 300000" leading-digit restriction.

   Step-by-Step Solution

   1. Digits available: $1,3,5,6,7,9$ (mod 4 these are $1,3,1,2,3,1$ respectively — note only the digit $6$ is $\equiv2\pmod4$ and none is $\equiv0\pmod4$).
   2. For last two digits (tens $a$, units $b$) to satisfy $10a+b\equiv2a+b\equiv0\pmod4$, since none of these digits is $\equiv0\pmod4$, only $b\equiv2\pmod4$ works — i.e. $b=6$ — and any $a\in\{1,3,5,7,9\}$ (each is $\equiv1$ or $3\pmod4$) satisfies $2a+6\equiv0\pmod4$ (check: $2(1)+6=8$, $2(3)+6=12$, all divisible by 4).
   3. So the number must end in $6$, and the tens digit can be any of $\{1,3,5,7,9\}$ — 5 choices — with the remaining 4 digits filling the first four places in $4!$ ways: $5\times4!=120$ six-digit multiples of 4 using these digits.
   4. Now impose "greater than $300000$": the leading digit must not be $1$ (a number starting with $1$ is below $300000$; starting with $3,5,7,9$ is automatically above). …

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3. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.The number of different nine digit numbers that can be formed by rearranging all the digits of the number $223355888$ so that odd digits always occupy even positions is (A) $180$ (B) $120$ (C) $60$ (D) $36$

   ›Reveal solutionSolution

   We count permutations of the multiset $\{2,2,3,3,5,5,8,8,8\}$ where odd digits (3,3,5,5) must occupy the four even positions (2nd,4th,6th,8th). The number of such arrangements is $1 \times 6 \times 6 = 36$, so the answer is (D).

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   Concept & Intuition

   The key is to treat positions and digits separately: first decide which digits go into the even slots, then arrange the remaining digits in the odd slots. Because the digits repeat, we must use the multinomial coefficient (permutations of a multiset) rather than simple factorials. The constraint “odd digits in even positions” fixes the type of digit for each slot, but not which specific odd digit goes where — so we count the ways to assign the four odd digits (two 3’s and two 5’s) to the four even positions, and independently arrange the five odd-position slots (which get the five even digits: two 2’s and three 8’s).

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   Step-by-step solution

   1. Identify positions and digit types

      A nine-digit number has positions 1 through 9 (from left to right). Even positions are 2, 4, 6, 8 — that’s 4 positions. Odd positions are 1, 3, 5, 7, 9 — that’s 5 positions.

      The digits available: two 2’s (even), two 3’s (odd), two 5’s (odd), three 8’s (even).

      So odd digits are {3,3,5,5} (four digits), even digits are {2,2,8,8,8} (five digits).

      The condition says: every even position must get an odd digit. Since there are exactly four odd digits and four even positions, this forces all odd digits into even positions, and all even digits into odd positions.

   2. Arrange the odd digits in the four even positions

      We have the multiset {3,3,5,5} to place in positions 2,4,6,8. The number of distinct arrangements is the multinomial coefficient:

   $\frac{4!}{2!\,2!} = \frac{24}{2 \times 2} = 6.$

   So there are 6 ways to fill the even positions.

   3. Arrange the even digits in the five odd positions The remaining digits are {2,2,8,8,8} for positions 1,3,5,7,9. The number of distinct arrangements is: $$$$

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4. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.If all the letters of the word RANKS are permutated in all possible ways and the words (with or without meaning) thus formed are arranged in dictionary order then the rank of the word RANKS is (A) $74$ (B) $76$ (C) $75$ (D) $77$

   ›Reveal solutionSolution

   The rank of a word in dictionary order is found by counting how many permutations would come before it. For RANKS, we fix letters one by one and sum the permutations of the remaining letters that start with a smaller letter. The rank is 75.

   We are asked: If all the letters of the word RANKS are permuted in all possible ways and the words (with or without meaning) thus formed are arranged in dictionary order, then the rank of the word RANKS is?

   Concept and Intuition: Permutations Without Repetition

   When we list all permutations of distinct letters in alphabetical order, we are essentially doing a "dictionary" ordering. The rank of a given word is simply 1 plus the number of words that come before it. To count those preceding words, we go letter by letter: for each position, we count how many permutations start with a letter smaller than the current letter (using the remaining letters), then fix that letter and move to the next position. This works because all letters are distinct — no repetitions to worry about.

   Let’s apply this to RANKS. First, sort the letters alphabetically: A, K, N, R, S.

   Now, step through the word RANKS.

   ---

   1. First letter: R

      Letters smaller than R in the sorted list: A, K, N.

      For each of these as the first letter, the remaining 4 letters can be arranged in $4! = 24$ ways.

      So words starting with A, K, or N: $3 \times 24 = 72$ words.

      After these, we fix R as the first letter and move to the second position.

   2. Second letter: A

      Remaining letters after fixing R: A, K, N, S. Sorted: A, K, N, S.

      Letters smaller than A? None. So 0 words start with RA and a smaller second letter.

      Fix A as the second letter. Remaining: K, N, S.

   3. Third letter: N

      Remaining letters sorted: K, N, S.

      Letters smaller than N: K. …

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5. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.All the letters of the word REMAIN are permuted in all possible ways and the words (with or without meaning) thus formed are arranged in the order as in the dictionary. The rank of the word REMAIN, when counted from the rank of the word MARINE beginning with the word MARINE itself, is (A) $266$ (B) $256$ (C) $272$ (D) $245$

   ›Reveal solutionSolution

   Computing dictionary ranks of MARINE (382) and REMAIN (637) among all permutations of {A,E,I,M,N,R}, the count from MARINE (inclusive) to REMAIN is 637-382+1 = 256.

   Concept and Intuition

   To rank a word among all permutations of its letters in dictionary order, fix the letters one position at a time, from left to right. At each position, count how many of the still-unused letters would come alphabetically before the letter actually chosen; each such letter, if placed there instead, would generate (number of remaining positions)! words that sort earlier. Summing these contributions and adding 1 gives the word's dictionary rank.

   Since REMAIN and MARINE are anagrams of each other (same multiset of 6 distinct letters {A,E,I,M,N,R}), both live within the same dictionary-ordered list of $6!=720$ permutations, and we can directly subtract their ranks.

   Step-by-Step Solution

   1. Alphabetical order of the letters: A, E, I, M, N, R.
   2. Rank of MARINE (letters M,A,R,I,N,E):
      • Position 1 = M: letters before M among {A,E,I,M,N,R} are A,E,I (3) → $3 \times 5! = 360$.
      • Position 2 = A: remaining letters {A,E,I,N,R}; none before A → $0 \times 4! = 0$.
      • Position 3 = R: remaining {E,I,N,R}; before R are E,I,N (3) → $3\times 3! = 18$.
      • Position 4 = I: remaining {E,I,N}; before I is E (1) → $1\times 2! = 2$.
      • Position 5 = N: remaining {E,N}; before N is E (1) → $1\times 1! = 1$.
      • Position 6 = E: remaining {E}; none before → 0.
      • Sum = 360+0+18+2+1+0 = 381. Rank = 381+1 = 382.
   3. Rank of REMAIN (letters R,E,M,A,I,N):
      • Position 1 = R: remaining {A,E,I,M,N,R}; before R are A,E,I,M,N (5) → $5\times 5! = 600$. …

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6. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.All possible 3 digit numbers are formed using the digits $0, 2, 3, 5, 7, 9$ without repeating any digit. Then the number of numbers among them which are divisible by 15 is (A) 11 (B) 14 (C) 16 (D) 36

   ›Reveal solutionSolution

   Divisibility by 15 = divisibility by 5 (last digit 0 or 5) AND by 3 (digit sum multiple of 3). Casework on the last digit and checking sums mod 3 gives 11 valid numbers.

   Concept and Intuition

   $15=3\times5$, and since $\gcd(3,5)=1$, a number is divisible by 15 exactly when it's divisible by both 3 and 5 simultaneously. Divisibility by 5 is a last-digit condition (must be 0 or 5), while divisibility by 3 is a digit-sum condition. So we fix the last digit to satisfy the divisible-by-5 rule, then among the remaining choices for the other two digits, keep only those whose total digit-sum (including the fixed last digit) is a multiple of 3 — and finally reject any arrangement that would put $0$ in the leading (hundreds) position.

   Step-by-Step Solution

   1. Digits mod 3: $0\to0,\ 2\to2,\ 3\to0,\ 5\to2,\ 7\to1,\ 9\to0$.

   Case A: last digit = 0. Need two more digits from $\{2,3,5,7,9\}$ whose sum is a multiple of 3 (since $0$ itself contributes nothing to the sum mod 3).

   • Pairs summing to $0 \pmod 3$: $\{3,9\}$ (both $\equiv0$), $\{7,2\}$ ($1+2\equiv0$), $\{7,5\}$ ($1+2\equiv0$) — 3 valid pairs.
   • Each pair fills the hundreds & tens slots in $2! = 2$ ways, and since none of $2,3,5,7,9$ is zero, both arrangements are valid (no leading-zero issue).
   • Numbers: $390,930,720,270,750,570$ — 6 numbers.

   Case B: last digit = 5. Need two more digits from $\{0,2,3,7,9\}$ such that (sum of the two) $+5 \equiv 0 \pmod 3$, i.e. sum of the two $\equiv 1 \pmod 3$ (since $5\equiv2$, need $2+\text{sum}\equiv0 \Rightarrow \text{sum}\equiv1$). …

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7. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.All possible 3 digit numbers are formed using all the digits $2, 3, 5, 7, 9$ without using any digit more than once. Among these 3 digit numbers, the number of numbers which are divisible by 3 but not divisible by 5 is (A) $24$ (B) $22$ (C) $20$ (D) $18$

   ›Reveal solutionSolution

   Only digit-triples whose sum is a multiple of $3$ give numbers divisible by $3$; subtract those ending in $5$. Answer $=24-4=20$.

   Divisibility by 3. A number is divisible by $3$ when its digit sum is. From $\{2,3,5,7,9\}$ choose $3$ distinct digits. Checking all $\binom{5}{3}=10$ triples, those with sum divisible by $3$ are:

   $\{2,3,7\}\;(12),\quad \{2,7,9\}\;(18),\quad \{3,5,7\}\;(15),\quad \{5,7,9\}\;(21).$

   That is $4$ triples. Each triple can be arranged in $3!=6$ ways, so numbers divisible by $3$:

   $4\times 6=24.$ …

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8. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.If all the letters of the word MESSI are permuted in all possible ways and the words [with or without meaning] thus formed are arranged in dictionary order, then the rank of the word MESSI is (A) $18$ (B) $27$ (C) $23$ (D) $26$

   ›Reveal solutionSolution

   This is a dictionary-rank problem with a repeated letter (S appears twice). Answer: rank $=27$.

   Concept and Intuition

   To rank a word in the dictionary ordering of all its letter-permutations, count how many permutations come strictly before it: for each position, count how many available letters are alphabetically smaller than the word's letter at that position, and for each such choice count the arrangements of the remaining letters (careful to divide by factorials of any repeated letters among the remaining set).

   Step-by-Step Solution

   1. Letters of MESSI, sorted alphabetically: $E, I, M, S, S$ (S repeated twice, so total distinct arrangements $= 5!/2! = 60$).
   2. First letter of MESSI is M. Letters smaller than M available: $E, I$.
      • Fix first letter = E: remaining letters $M,S,S,I$ arrange in $4!/2! = 12$ ways.
      • Fix first letter = I: remaining letters $M,E,S,S$ arrange in $4!/2! = 12$ ways.
      • Total words before any M-word: $12+12 = 24$.
   3. Now consider words starting with M. Remaining letters to place: $E, I, S, S$ (alphabetical order $E<I<S$).
      • Second letter of MESSI is E, the smallest available, so no words with second letter I or S come before it.
   4. Within "M,E,,,_", remaining letters are $I, S, S$; alphabetical order $I<S<S$. …

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9. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.The number of all five letter words (with or without meaning) having atleast one repeated letter that can be formed by using the letters of the word INCONVENIENCE is (A) $2025$ (B) $2765$ (C) $3265$ (D) $3205$

   ›Reveal solutionSolution

   Summing all repeated-letter cases gives $3605$ five-letter words with at least one repeated letter — option (D).

   Concept and Intuition

   INCONVENIENCE has letter frequencies N:4, E:3, I:2, C:2, O:1, V:1 (6 distinct letters). 'At least one repeated letter' is best counted by adding every arrangement pattern that uses a repeat, i.e. all patterns except the all-distinct one.

   Step-by-Step Solution

   1. One pair + 3 distinct (2+1+1+1): pair from {N,E,I,C}=4; choose 3 of remaining 5 distinct =$\binom{5}{3}=10$; arrangements $\frac{5!}{2!}=60$ → $4\times10\times60=2400$.
   2. Two pairs + 1 (2+2+1): pairs $\binom{4}{2}=6$; single from remaining 4 =4; arrangements $\frac{5!}{2!2!}=30$ → $6\times4\times30=720$.
   3. One triple + 2 distinct (3+1+1): triple from {N,E}=2; 2 of remaining 5 =$\binom{5}{2}=10$; arrangements $\frac{5!}{3!}=20$ → $2\times10\times20=400$.
   4. Triple + pair (3+2): triple from {N,E}=2; pair from remaining {N,E,I,C}=3; arrangements $\frac{5!}{3!2!}=10$ → $2\times3\times10=60$. …

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10. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
    Q.The number of ways of arranging all the letters of the word PERFECTION such that there must be exactly two consonants between any two vowels is (A) $4! + 6!$ (B) $3! + 6!$ (C) $2! \, 3! \, 6!$ (D) $\dfrac{6!}{4!}$

    ›Reveal solutionSolution

    With exactly 6 distinct consonants and 4 vowels (one repeated), the 'exactly two consonants between consecutive vowels' condition forces one single pattern V-CC-V-CC-V-CC-V; multiplying the vowel- and consonant-arrangement counts gives $2!\,3!\,6!$.

    Concept and Intuition

    PERFECTION splits into 4 vowels (E, E, I, O) and 6 consonants (P, R, F, C, T, N — all different). A '2 consonants between each pair of consecutive vowels' constraint with exactly 4 vowels creates exactly 3 gaps between them, needing $3\times2=6$ consonants total — which is exactly the number of consonants available. That means there's no room for any consonant before the first vowel or after the last one: the whole word is forced into one rigid template, and counting arrangements becomes a simple product of two independent permutation counts (vowels among themselves, consonants among themselves).

    Step-by-Step Solution

    1. List the letters of PERFECTION: P, E, R, F, E, C, T, I, O, N — 10 letters total.
    2. Vowels: E, E, I, O — 4 vowels, with E repeated twice.
    3. Consonants: P, R, F, C, T, N — 6 consonants, all distinct.
    4. With 4 vowels placed in the word, there are exactly 3 gaps between consecutive vowels (between vowel 1 & 2, vowel 2 & 3, vowel 3 & 4). The condition demands exactly 2 consonants in each such gap: $3\times2=6$ consonants needed — which is precisely all 6 available consonants.
    5. Since all 6 consonants are used up filling the 3 internal gaps, there are no consonants left to place before the first vowel or after the last vowel. So the entire arrangement must follow the single fixed template: $V\ C\,C\ V\ C\,C\ V\ C\,C\ V$ (4 vowel-slots and 6 consonant-slots, in this exact fixed skeleton).
    6. Fill the vowel slots: arrange E, E, I, O (4 letters, one pair identical) in the 4 designated vowel slots: $\dfrac{4!}{2!}=12$ ways. Note $12=3!\times2!$ (since $3!=6,\ 2!=2,\ 6\times2=12$). …

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11. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
    Q.If all the letters of the word COMBINATION are arranged in all possible ways to form 11 letter words (with or without meaning), then the number of words among them in which C and N occupy the end positions and no vowel appears exactly in the middle position is (A) $\dfrac{5}{2}(8!)$ (B) $4(8!)$ (C) $2(8!)$ (D) $36(7!)$

    ›Reveal solutionSolution

    This tests constrained permutations with repeated letters: fix the ends, then forbid a vowel at the middle. Answer: $2(8!)$.

    Concept and Intuition

    COMBINATION $=$ C,O,M,B,I,N,A,T,I,O,N — 11 letters with repeats O$\times2$, I$\times2$, N$\times2$. When a problem fixes certain letters at certain positions, the standard technique is: place the fixed letters first (counting the ways to do that), then permute the remaining multiset freely in the remaining positions, applying any leftover restriction (here, "no vowel in the middle") as a restricted count within that remaining arrangement.

    Step-by-Step Solution

    1. Letters: C(1), O(2), M(1), B(1), I(2), N(2), A(1), T(1) — total 11, vowels $=$ O,O,I,I,A (5), consonants $=$ C,M,B,N,N,T (6).
    2. "C and N occupy the end positions" (positions 1 and 11): either C is at position 1 and an N at 11, or an N is at position 1 and C at 11 — 2 ways to assign which end gets which letter (using one of the two identical N's; using either copy of N gives the same word, so there's no double count).
    3. After placing C and one N at the ends, the remaining 9 letters to fill positions 2–10 are: O,O,I,I,M,B,N,A,T (vowels O,O,I,I,A = 5; consonants M,B,N,T = 4, each single).
    4. Position 6 (the true middle of 11 positions) is one of these 9 slots, and it must NOT be a vowel, i.e., it must be one of the 4 single consonants M,B,N,T. …

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12. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
    Q.If four letters are chosen from the letters of the word ASSIGNMENT and are arranged in all possible ways to form 4 letter words (with or without meaning), then total number of such words that can be formed is (A) 1680 (B) 2184 (C) 2196 (D) 2190

    ›Reveal solutionSolution

    Break the word ASSIGNMENT's letters into distinct singles and repeated letters, count 4-letter selections case-by-case (all distinct / one pair / two pairs), and sum the arrangements. Answer: $2190$.

    Concept and Intuition

    When letters repeat, "choose then arrange" must be split into cases based on how many repeated letters are used, since the arrangement count for a multiset (with repeats) differs from that of all-distinct letters ($4!/2!$ vs $4!$).

    Step-by-Step Solution

    1. ASSIGNMENT = A, S, S, I, G, N, M, E, N, T — 10 letters total. Distinct letter types: A, S, I, G, N, M, E, T (8 types); S and N each occur twice, the rest occur once.
    2. Case 1 — all 4 chosen letters distinct types: choose 4 of the 8 distinct types: $\binom{8}{4}=70$. Each such set of 4 distinct letters arranges in $4!=24$ ways. Subtotal $=70\times24=1680$. …

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