Permutation and combination practice · Mixed difficulty

Option A (93324) is correct because it calculates the sum of all possible numbers formed by arranging the digits 2, 3, 4, and 5.

Option D (1630) is correct because the total number of positive integers formed is the sum of one-digit to six-digit numbers, accounting for the restriction on using each digit more than once and excluding zero as the first digit.

Option C ($16 \times 1111 \times 3$ !) is correct because it calculates the sum of all numbers formed by the digits 1, 3, 5, 7 without repetition.

Option A (900) is correct because it subtracts arrangements with two R's together from the total arrangements of the word ARRANGE.

Option B (450000) is correct because it accounts for the possibilities of odd and even sums of the first five digits and then chooses the last digit accordingly.

Option A (6) is correct because ${ }^{12} {P}_{{r}}={ }^{11} {P}_6+6 .{ }^{11} {P}_5$ simplifies to ${ }^{12}P_r = { }^{12}P_6$, thus $r=6$.

Option A (5199960) is correct because it calculates the sum of all numbers greater than 10000 formed by the digits 0, 2, 4, 6, and 8 without repetition.

Option D (720) is correct because there are 3 choices for the first even digit, 2 for the last, and then 5P4 ways to arrange the remaining digits.

The correct answer is C (360) because it represents the number of 5-digit numbers of the form abcba where a < b.

Option B (55) is correct because it represents the number of points satisfying the given conditions, which translates to finding positive integral solutions of a related equation.

Option C (240) is correct because it correctly calculates the number of arrangements where A, I, and O are only in odd places, considering the repetitions of I and N.

Option A (${\;^{56}}{C_4}$) is correct because the given summation can be simplified using the identity ${\;^n}{C_r} + {\;^n}{C_{r - 1}} = {\;^{n + 1}}{C_r}$ repeatedly.

Option A (9360) is correct because it is the sum of all divisors, calculated by multiplying the sums of powers of each prime factor.

Option D (19 !- 18 ! 2 !) is correct because it subtracts the number of arrangements where the two delegates are together (18! * 2!) from the total number of arrangements (19!).

The correct answer is D (58th) because the rank of the word SMALL is determined by arranging all possible words in dictionary order.

Option D (${ }^{r} P_{3} \times{ }^{n-3} P_{r-3}$) is correct because it accounts for selecting and arranging the remaining objects after including the 3 particular ones.

Option C (24) is correct because it represents the highest power of 5 that divides 100!, which determines the number of trailing zeros.

Option A (120) is correct because it represents the total number of selections with at least one boy.

Option B (190) is correct because the total number of handshakes is calculated using combinations, specifically ${}^{20}C_2$, which equals 190.

Option D (360) is correct because it treats the two R's as one unit and then arranges the remaining letters, accounting for the repeated A.

Option A (3265920) is correct because we calculate the number of 8-digit numbers with distinct digits, considering that the first digit cannot be 0.

Option A (P(n-p, r)) is correct because it represents the number of permutations when p particular things are excluded from n different things, and r items are taken at a time.

Option B ($2! \times 18!$) is correct because the host and the two persons can be considered as a single unit, and the two persons can be arranged in 2! ways.

Option D (${ }^{4 n} C_{2 n}$) is the correct answer because $2^n E$ can be expressed as ${ }^{4 n} C_{2 n} \cdot n !$, which is divisible by ${ }^{4 n} C_{2 n}$.

Option A (2475) is correct because it is the number of pairs (m, n) where the digits of n are less than or equal to the corresponding digits of m, calculated as the product of sums.

Option C (41) is correct because it involves solving an equation derived from the ratio of two permutation formulas.

Option B (${3^{.n + 1}}{C_4}$) is correct because by substituting $m = {\;^n}{C_2}$ into ${\^m}{C_2}$ and simplifying, we arrive at $3{\ ^{n + 1}}{C_4}$.

Option B ($\frac{5 !}{2 ! 2 !}$) is correct because it accounts for the repetitions of the digits 2 and 3.

The correct answer is D (8) because solving the equation derived from the given condition $T_{n+1} - T_n = 28$ yields n = 8.

Option B (8) is correct because it is the least positive integral value of x that satisfies the given inequality.