Permutation and combination practice · Medium

Option A ($\frac{{10!}}{6}$) is correct because A, B, and C can be arranged in 3! = 6 ways, but only one of these ways (A before B before C) is desired, so we divide the total permutations by 6.

Option B (70) is correct because it applies Burnside's Lemma or Polya's Enumeration Theorem to count the distinct colorings of the vertices of a square under rotations, while keeping reflections distinct.

Option C (48) is correct because the coefficient of the quadratic term cannot be zero, while the other coefficients can be any of the four given values.

Option C (300) is correct because we count the occurrences of 5 in the units, tens, and hundreds places from 1 to 999.

Option C (101) is correct because it calculates the number of words preceding 'WATER' in the dictionary by considering the alphabetical order of the letters.

Option A ($2 \times 12!$) is correct because the Secretary and Deputy Secretary are fixed relative to the Chairman, and the remaining members are arranged in a circle.

Option D ($\frac{{4\;!3\;!}}{2}$) is correct because it accounts for arranging the consonants and vowels separately, considering the repetition of the vowel 'A'.

Option A ($(n+1) !$) is correct because the sum of the series can be simplified to a telescoping sum, resulting in (n+1)!.

Option A (${ }^{n-1} C_2$) is correct because it represents the number of positive integral solutions for the given equation using stars and bars method.

Option C (84) is correct because it accounts for all numbers greater than 24000 formed without repetition.

Option C (144) is correct because it accounts for the positions of R and D with two letters in between, the arrangements of those two letters, and the arrangements of R and D themselves.

Option B (240) is correct because it subtracts the number of words where B and H are together from the total number of words.

Option A (35) is correct because the '+' signs create 7 spaces where the '-' signs can be placed, and we need to choose 4 of those spaces.

Option C (96) is correct because it counts all the words that would appear before "COCHIN" in an alphabetical listing.

Option A (666) is correct because it represents the number of integers between 2 and 2000 that are not divisible by 2 or 3, calculated using the principle of inclusion-exclusion.

Option A (FATHER) is correct because when the words are arranged alphabetically, FATHER is the 261st word.

Option B (4960) is correct because it subtracts the number of four-digit numbers without repetition from the total number of four-digit numbers.

The correct answer is A (24) because it considers arranging U, CC, and E, and then placing the three S's in the available spaces such that no two S's are together.

Option C (${ }^{17} {C}_7$) is correct because it represents the number of ways to distribute identical objects among persons without any restrictions using stars and bars method.

Option D (91) is correct because it is the total number of possible outcomes minus the number of outcomes where no dice shows 5.

Option B (35) is correct because it represents the number of ways to distribute 16 identical items among 4 people, with each person receiving at least 3 items.

Option B (60) is correct because each digit appears in the unit's place $3!$ times, and the sum of the digits is $1+2+3+4=10$, so the sum of digits in the unit's place is $3! \times 10 = 60$.

Option A (264) is correct because it represents the number of ways to place 6 letters into wrong envelopes, considering 5 letters are misplaced.

Option A (46) is correct because it represents the number of proper divisors of 2520.

Option B (190) is correct because it sums the number of integral points on each line parallel to the y-axis within the triangle's interior.

Option A (150) is correct because the number of arrangements is calculated by considering the possible arrangements with at least one 'b' between two 'a's and then subtracting from the total arrangements.

Option B (${ }^{n-2} C_3$) is correct because it represents the number of ways to select three non-adjacent objects from n objects arranged in a row.

Option A (4) is correct because solving the given ratio of permutations leads to n = 4.

Option B (6) is correct because after calculating the sum of factorials, the tens digit is determined to be 6.

Option A (56) is correct because it represents the total number of 6-digit numbers that can be formed where the sum of their digits is 51. The possible combinations are derived by considering the minimum digit value of 6.