Question 1
How many different words from the letters of the word MATHEMATICS can be formed so that all the vowels always come together in any word?
Explanation
MATHEMATICS has 11 letters: M(2), A(2), T(2), H, E, I, C, S. The vowels are A, A, E, I (4 vowels) and the consonants are M, M, T, T, H, C, S (7 consonants, with M and T each repeated).
Treat all 4 vowels as one single block. This gives 8 units to arrange (7 consonants + 1 vowel-block), with M and T each repeated twice:
Arrangements of these 8 units = 8!2! × 2! = 403204 = 10,080.
Within the vowel block, the 4 vowels A, A, E, I can be arranged among themselves in 4!2! = 12 ways (dividing by 2! for the repeated A).
Total words = 10,080 × 12 = 1,20,960.
Treat all 4 vowels as one single block. This gives 8 units to arrange (7 consonants + 1 vowel-block), with M and T each repeated twice:
Arrangements of these 8 units = 8!2! × 2! = 403204 = 10,080.
Within the vowel block, the 4 vowels A, A, E, I can be arranged among themselves in 4!2! = 12 ways (dividing by 2! for the repeated A).
Total words = 10,080 × 12 = 1,20,960.
Question 2
Find n if nP5 =20 nP3
Explanation
nP5nP3 = (n-3)!(n-5)! = (n-3)(n-4).
Setting this equal to 20: (n-3)(n-4) = 20 ⇒ n2 - 7n + 12 = 20 ⇒ n2 - 7n - 8 = 0 ⇒ (n-8)(n+1) = 0.
Since n must be a positive integer, n = 8.
Setting this equal to 20: (n-3)(n-4) = 20 ⇒ n2 - 7n + 12 = 20 ⇒ n2 - 7n - 8 = 0 ⇒ (n-8)(n+1) = 0.
Since n must be a positive integer, n = 8.
Question 3
In a school, for a class monitor selection, there are 6 candidates, and students need to choose up to 3 monitors. A student can vote for 1 or 2 or 3 candidates. In how many ways a student can vote?
Explanation
The student can select a group of exactly 1, exactly 2, or exactly 3 candidates out of 6, so the total number of ways is the sum of the combinations for each case:
Ways to choose 1 = 6C1 = 6
Ways to choose 2 = 6C2 = 15
Ways to choose 3 = 6C3 = 20
Total = 6 + 15 + 20 = 41.
Ways to choose 1 = 6C1 = 6
Ways to choose 2 = 6C2 = 15
Ways to choose 3 = 6C3 = 20
Total = 6 + 15 + 20 = 41.
Question 4
Different words are made with rearrangement of letters of the word "TROPICAL" in a way that the vowels occupy odd places when counted from left. How many such words are there?
Explanation
TROPICAL has 8 distinct letters: T, R, O, P, I, C, A, L. Its vowels are O, I, A (3 vowels), and its consonants are T, R, P, C, L (5 consonants).
Out of the 8 positions, the odd positions (counted from the left) are 1, 3, 5, 7 -- 4 odd positions in total, but only 3 vowels need to be placed. So first choose 3 of these 4 odd positions for the vowels: 4C3 = 4 ways, and arrange the 3 distinct vowels in them: 3! = 6 ways.
The remaining 5 positions (1 leftover odd position + all 4 even positions) are filled by the 5 distinct consonants: 5! = 120 ways.
Total words = 4 × 6 × 120 = 2,880.
Out of the 8 positions, the odd positions (counted from the left) are 1, 3, 5, 7 -- 4 odd positions in total, but only 3 vowels need to be placed. So first choose 3 of these 4 odd positions for the vowels: 4C3 = 4 ways, and arrange the 3 distinct vowels in them: 3! = 6 ways.
The remaining 5 positions (1 leftover odd position + all 4 even positions) are filled by the 5 distinct consonants: 5! = 120 ways.
Total words = 4 × 6 × 120 = 2,880.
Question 5
If nCr-1 = 28, nCr = 56, nCr+1 = 70, then the value of n and r are:
Explanation
Using the ratio nCrnCr-1 = n - r + 1r: 5628 = 2 = n - r + 1r, so n - r + 1 = 2r, i.e. n = 3r - 1.
Similarly, nCr+1nCr = n - rr + 1: 7056 = 1.25 = n - rr + 1, so n - r = 1.25(r + 1).
Substituting n = 3r - 1: 3r - 1 - r = 1.25r + 1.25 ⇒ 2r - 1 = 1.25r + 1.25 ⇒ 0.75r = 2.25 ⇒ r = 3.
Then n = 3(3) - 1 = 8. So n = 8, r = 3.
Similarly, nCr+1nCr = n - rr + 1: 7056 = 1.25 = n - rr + 1, so n - r = 1.25(r + 1).
Substituting n = 3r - 1: 3r - 1 - r = 1.25r + 1.25 ⇒ 2r - 1 = 1.25r + 1.25 ⇒ 0.75r = 2.25 ⇒ r = 3.
Then n = 3(3) - 1 = 8. So n = 8, r = 3.
Question 6
In a meeting, 5 analysts, 2 consultants, and 3 managers are to be seated in a row. If members of the same profession must sit together, in how many ways can they be seated?
Explanation
Treat each profession group as a single block: 3 blocks can be arranged among themselves in 3! ways.
Within each block, members can be arranged internally: analysts in 5! ways, consultants in 2! ways, managers in 3! ways.
Total arrangements = 3! × 5! × 2! × 3! = 6 × 120 × 2 × 6 = 8,640.
Within each block, members can be arranged internally: analysts in 5! ways, consultants in 2! ways, managers in 3! ways.
Total arrangements = 3! × 5! × 2! × 3! = 6 × 120 × 2 × 6 = 8,640.
Question 7
The total numbers greater than 2000 that can be formed with the digits 1, 2, 3, 4, 5 with no digit being repeated in any number are:
Explanation
Numbers greater than 2000 can have 4 digits (starting with 2, 3, 4 or 5) or 5 digits (any arrangement of all 5 digits automatically exceeds 2000).
4-digit numbers: first digit has 4 choices (2, 3, 4 or 5), remaining 3 digits arranged from the remaining 4 digits in 4P3 = 24 ways, giving 4 × 24 = 96 numbers.
5-digit numbers: all 5 digits arranged in 5! = 120 ways.
Total = 96 + 120 = 216.
4-digit numbers: first digit has 4 choices (2, 3, 4 or 5), remaining 3 digits arranged from the remaining 4 digits in 4P3 = 24 ways, giving 4 × 24 = 96 numbers.
5-digit numbers: all 5 digits arranged in 5! = 120 ways.
Total = 96 + 120 = 216.
Question 8
How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9 which are divisible by 5 and none of the digits is repeated?
Explanation
For a number to be divisible by 5, its units digit must be 5 (the only multiple of 5 among the given digits), so the units place is fixed.
The remaining two places (hundreds and tens) are filled from the remaining 5 digits without repetition: 5P2 = 5 × 4 = 20.
Total 3-digit numbers = 20.
The remaining two places (hundreds and tens) are filled from the remaining 5 digits without repetition: 5P2 = 5 × 4 = 20.
Total 3-digit numbers = 20.
Question 9
Find out the number of 5-digit even numbers that can be formed from digits 1 to 7 without repetition of any digit.
Explanation
For the number to be even, the units digit must be one of {2, 4, 6}: 3 choices.
The remaining 4 positions are filled from the remaining 6 digits without repetition: 6P4 = 6 × 5 × 4 × 3 = 360.
Total = 3 × 360 = 1,080.
The remaining 4 positions are filled from the remaining 6 digits without repetition: 6P4 = 6 × 5 × 4 × 3 = 360.
Total = 3 × 360 = 1,080.
Question 10
In how many ways can 5 Indians and 5 Americans be seated around a table so that no two Indians are in adjacent positions?
Explanation
First seat the 5 Americans around the circular table: (5-1)! = 4! ways.
This creates 5 gaps between adjacent Americans. To ensure no two Indians sit together, place one Indian in each of these 5 gaps: 5! ways.
Total arrangements = 4! × 5!.
This creates 5 gaps between adjacent Americans. To ensure no two Indians sit together, place one Indian in each of these 5 gaps: 5! ways.
Total arrangements = 4! × 5!.
Question 11
A group consists of 7 men and 5 women. In how many ways can a group of 4 members be selected if the group has no women?
Explanation
If no women are included, all 4 members must be chosen from the 7 men.
Number of ways = 7C4 = 7!4!3! = 35.
Number of ways = 7C4 = 7!4!3! = 35.
Question 12
Four cards are drawn at random from a standard deck of 52 playing cards without replacement. In how many ways can it be done such that the selected cards consist of exactly one Jack and three Aces?
Explanation
There are exactly 4 Jacks and 4 Aces in a standard deck.
Number of ways to choose exactly 1 Jack from 4: 4C1 = 4.
Number of ways to choose exactly 3 Aces from 4: 4C3 = 4.
Total ways = 4 × 4 = 16.
Number of ways to choose exactly 1 Jack from 4: 4C1 = 4.
Number of ways to choose exactly 3 Aces from 4: 4C3 = 4.
Total ways = 4 × 4 = 16.