Permutation and combination practice · Medium

Option C (13) is correct because it represents the number of possible values for the ordered triplet (a, b, c) when a, b, and c are in AP and their sum is 21.

The correct answer is 36 because there are 3! ways to arrange the consonants in the odd places and 3! ways to arrange the vowels in the even places.

Option C (78) is correct because it accounts for the combinations of selecting 2 consonants and 2 vowels, considering repetitions.

Option A (325) is correct because it sums the number of words of length 1 to 5 that can be formed from the letters of 'KUBER' without repetition.

Option D (-12) is correct because after simplifying the expression for E, it becomes $2^{-36}$ which is equivalent to $8^{-12}$.

Option C ([-3,-2 $\sqrt{2}$) ∪(2 $\sqrt{2}$, 3]) is correct because it correctly determines the range of k based on the given condition and the properties of combinations.

Option A (601) is correct because it calculates the rank of the word SACHIN based on dictionary order.

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

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.

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

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 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 (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 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 ${ }^{4 n} C_{2 n}$.

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 A ($2^8-2$) is correct because the given expression can be simplified using binomial coefficient identities to equal $2^8-2$.

Option C (35) is correct because the number of ways to select four non-adjacent persons from ten is equivalent to arranging 4 ones and 6 zeros with no two ones adjacent, which can be done in ${ }^7 C_4 = 35$ ways.

Option C (60) is correct because it accurately counts the number of ways to distribute the toys such that each boy gets at least one and no two boys get the same number of toys.

Option C (NAAIG) is correct because when arranging the permutations in dictionary order, NAAIG is the 50th word.

Option D (89) is correct because it represents the rank of the word TOUGH when the letters of GHOTU are arranged alphabetically.

Option C (89) is correct because it represents the correct lexicographical order of the word TOUGH among all permutations of OUGHT.

Option B (300) is correct because it accurately counts the occurrences of the digit 3 in the integers from 1 to 1000.

Option B (14400) is correct because it calculates the number of arrangements of the letters in FRACTION such that no two vowels are together.

Option B (90) is correct because it is the total number of functions satisfying the given condition, calculated by considering all possible cases.

The correct answer is 374 because the number of integers between 1000 and 4000 that can be formed using the digits 0, 1, 2, 3, 4 with repetition allowed is 374.

Option D (None of the above) is incorrect. Option A (196) is correct because it calculates the number of ways to choose a committee of 5 from 10 candidates, including the oldest and excluding the youngest.

Option D (72) is correct because there are 48 numbers excluding 3 and 24 numbers excluding 6 that are divisible by 6.