Question 1

Passage: A 4-digit number is selected at random formed by using the digits 0, 1, 2, 3 and 4 (where repetition of digits is not allowed). Question: What is the probability that the number selected is d

Accepted Answer

Total number of 4-digit numbers formed using 0, 1, 2, 3, 4 without repetition: The thousands place can be filled in 4 ways (excluding 0). The remaining 3 places can be filled in ^ 4 P_ 3 = 24 ways. Total numbers = 4 24 = 96. For the number to be divisible by 2, the units digit must be 0, 2, or 4. Case 1: Units digit is 0. The remaining 3 places can be filled in ^ 4 P_ 3 = 24 ways. Case 2: Units digit is 2 or 4. The units place can be filled in 2 ways. The thousands place can be filled in 3 ways

Question 2

How many numbers greater than 1000 can be formed using the digits 0, 1, 2 and 3 (repetition of digits is not allowed) ?

Accepted Answer

To form a number greater than 1000 using the digits 0, 1, 2, 3 without repetition, the number must be a 4-digit number. The thousands place cannot be 0, so it can be filled in 3 ways (using 1, 2, or 3). The remaining 3 places can be filled with the remaining 3 digits in 3! ways. Total number of such numbers = 3 3! = 3 6 = 18. Answer: 18

Question 3

What is the maximum number of points of intersection of 5 circles ?

Accepted Answer

Two circles can intersect at a maximum of 2 distinct points. To find the maximum number of points of intersection of 5 circles, we select any 2 circles from the given 5 circles and multiply the number of pairs by 2. The number of ways to select 2 circles from 5 is ^ 5 C_ 2 . Maximum number of points of intersection = 2 ^ 5 C_ 2 = 2 5 4 2 1 = 20 Answer: 20

Question 4

Passage: A 4-digit number is selected at random formed by using the digits 0, 1, 2, 3 and 4 (where repetition of digits is not allowed). Question: What is the probability that the number selected is d

Accepted Answer

Total number of 4-digit numbers formed using 0, 1, 2, 3, 4 without repetition: The thousands place can be filled in 4 ways (excluding 0). The remaining 3 places can be filled in ^ 4 P_ 3 = 24 ways. Total possible numbers = 4 24 = 96. A number is divisible by 3 if the sum of its digits is a multiple of 3. The sum of all given digits is 0 + 1 + 2 + 3 + 4 = 10. To choose 4 digits whose sum is a multiple of 3, the excluded digit must leave a remainder of 1 when divided by 3. Thus, the excluded digit

Question 5

Passage: A 4-digit number is selected at random formed by using the digits 0, 1, 2, 3 and 4 (where repetition of digits is not allowed). Question: What is the probability that the number selected is d

Accepted Answer

Total number of 4-digit numbers formed using 0, 1, 2, 3, 4 without repetition: The thousands place can be filled in 4 ways (excluding 0). The remaining 3 places can be filled in ^ 4 P_ 3 = 24 ways. Total possible numbers = 4 24 = 96. A number is divisible by 4 if the number formed by its last two digits is divisible by 4. The possible last two digits from the given set are 04, 12, 20, 24, 32, 40. Case 1: Last two digits are 04, 20, or 40 (3 cases). Since 0 is used, the first two places can be fi

Question 6

If n= ^ m C_2 then what is ^ n C_2 equal to ?

Accepted Answer

Given n = ^ m C_ 2 = m(m-1) 2 Evaluating ^ n C_ 2 = n(n-1) 2 Substituting the value of n: ^ n C_ 2 = m(m-1) 2 ( m(m-1) 2 - 1 ) 2 ^ n C_ 2 = m(m-1)(m^2 - m - 2) 8 Factorizing m^2 - m - 2 = (m-2)(m+1): ^ n C_ 2 = (m+1)m(m-1)(m-2) 8 Multiplying and dividing by 3: ^ n C_ 2 = 3 (m+1)m(m-1)(m-2) 24 ^ n C_ 2 = 3 ^ m+1 C_ 4 Answer: 3 ^ m+1 C_4

Question 7

Passage: A 4-digit number is selected at random formed by using the digits 0, 1, 2, 3 and 4 (where repetition of digits is not allowed). Question: What is the probability that the number selected does

Accepted Answer

The total number of 4-digit numbers formed using the digits 0, 1, 2, 3, 4 without repetition is calculated by filling the thousands place in 4 ways (excluding 0) and the remaining 3 places in ^ 4 P_ 3 ways. Total outcomes = 4 4 3 2 = 96 The number of 4-digit numbers formed without using 0 at any position means using only the digits 1, 2, 3, 4. Favorable outcomes = ^ 4 P_ 4 = 4 3 2 1 = 24 Required probability = 24 96 = 1 4 = 0.25 Answer: 0 25

Question 8

How many 4-digit numbers that are divisible by 4 can be formed using the digits 1, 2, 3 and 4 (repetition of digits is not allowed)?

Accepted Answer

A number is divisible by 4 if the number formed by its last two digits is divisible by 4. Using the digits 1, 2, 3, 4 without repetition, the possible two-digit numbers at the end that are divisible by 4 are 12, 24, and 32. For each of these endings, the remaining two digits can be arranged in the first two places in 2! = 2 ways. Total number of such 4-digit numbers is 3 2 = 6. Answer: 6

Question 9

For the following two (02) items: There are 8 points on a plane out of which 4 points are collinear. How many quadrilaterals can be formed by joining these points?

Accepted Answer

Total number of points is 8. Number of collinear points is 4. Total number of ways to select 4 points from 8 points is ^ 8 C_ 4 = 70. A quadrilateral cannot be formed if 3 or 4 selected points are collinear. Number of ways to select 4 collinear points is ^ 4 C_ 4 = 1. Number of ways to select 3 collinear points and 1 non-collinear point is ^ 4 C_ 3 ^ 4 C_ 1 = 4 4 = 16. Total number of quadrilaterals that can be formed is 70 - (1 + 16) = 53. Answer: 53

Question 10

Let y = x! and z = (2x)!. If (z/y) = 120, then what is the value of (3x)!?

Accepted Answer

Given y = x! and z = (2x)!. We are given z y = 120. Substituting the values of y and z, we get: (2x)! x! = 120 Checking for small integer values of x: For x = 1, 2! 1! = 2 For x = 2, 4! 2! = 24 2 = 12 For x = 3, 6! 3! = 720 6 = 120 Thus, x = 3. We need to find the value of (3x)!. Substituting x = 3, we get (3 3)! = 9!. 9! = 9 8 7 6 5 4 3 2 1 = 362880. Answer: 362880

Question 11

In how many ways can the letters of the word DELHI be arranged keeping the positions of vowels and consonants unchanged?

Accepted Answer

The word DELHI consists of 5 letters: 3 consonants (D, L, H) and 2 vowels (E, I). The arrangement of letters in the word is Consonant-Vowel-Consonant-Consonant-Vowel. Since the positions of vowels and consonants must remain unchanged, the consonants can only be placed in the 1st, 3rd, and 4th positions, and the vowels can only be placed in the 2nd and 5th positions. The 3 consonants can be arranged in their 3 positions in 3! ways. The 2 vowels can be arranged in their 2 positions in 2! ways. Tot

Question 12

How many sides are there in a polygon which has 20 diagonals?

Accepted Answer

The number of diagonals of an n-sided polygon is given by ^ n C_ 2 - n = n(n-3) 2 . Given that the number of diagonals is 20. n(n-3) 2 = 20 n^2 - 3n - 40 = 0 (n-8)(n+5) = 0 Since the number of sides n must be a positive integer, n = 8. Answer: 8

Question 13

How many 7-letter words (with or without meaning) can be constructed using all the letters of the word CAPITAL so that all consonants come together in each word?

Accepted Answer

The word CAPITAL has 7 letters. Vowels: A, I, A (3 letters, where A is repeated twice) Consonants: C, P, T, L (4 distinct letters) Since all consonants must come together, they are treated as a single unit: (C, P, T, L). Now, there are 4 units to arrange: the consonant group and the 3 vowels (A, I, A). The number of ways to arrange these 4 units is 4! 2! because the letter A appears twice. 4! 2! = 12 The 4 consonants can be arranged among themselves in 4! ways. 4! = 24 Total number of words = 12

Question 14

What is the number of selections of at most 3 things from 6 different things?

Accepted Answer

The number of ways to select r things from n different things is given by ^ n C_ r . The phrase "at most 3 things" means we can select 0, 1, 2, or 3 things from the 6 available different things. The total number of selections is the sum of these possibilities: ^ 6 C_ 0 + ^ 6 C_ 1 + ^ 6 C_ 2 + ^ 6 C_ 3 Calculating each term: ^ 6 C_ 0 = 1 ^ 6 C_ 1 = 6 ^ 6 C_ 2 = 6 5 2 1 = 15 ^ 6 C_ 3 = 6 5 4 3 2 1 = 20 Total number of selections = 1 + 6 + 15 + 20 = 42. Answer: 42

Question 15

For the following two (02) items: There are 8 points on a plane out of which 4 points are collinear. How many triangles can be formed by joining these points?

Accepted Answer

Total number of points on the plane is n = 8. Number of collinear points is m = 4. The total number of triangles that can be formed by joining n points out of which m are collinear is given by ^ n C_ 3 - ^ m C_ 3 . Substituting the values, we get ^ 8 C_ 3 - ^ 4 C_ 3 . ^ 8 C_ 3 = 8 7 6 3 2 1 = 56 ^ 4 C_ 3 = 4 3 2 3 2 1 = 4 Total number of triangles = 56 - 4 = 52. Answer: 52

Question 16

Let n be a natural number. The number of consecutive zeros at the end of the expansion of n! is exactly 2. How many values of n are possible?

Accepted Answer

The number of consecutive zeros at the end of n! is determined by the exponent of 5 in the prime factorization of n!. Using Legendre's formula, the exponent of 5 in n! is given by: E_5(n!) = n 5 + n 5^2 + n 5^3 + Given that E_5(n!) = 2, n must be less than 25, which means n 25 = 0 and all higher terms are also 0. Thus, the equation simplifies to: n 5 = 2 This inequality holds for: 2 n 5 10 n Since n is a natural number, the possible values of n are 10, 11, 12, 13, 14. There are exactly 5 possibl

Question 17

If the number of selections of r as well as (n+r) things from 5n different things are equal, then what is the value of r?

Accepted Answer

The number of selections of r things from 5n different things is given by ^ 5n C_ r . The number of selections of (n+r) things from 5n different things is given by ^ 5n C_ n+r . Given that ^ 5n C_ r = ^ 5n C_ n+r . Using the property of combinations, if ^ N C_ x = ^ N C_ y , then either x = y or x + y = N. Since r n+r (for n 0), we must have: r + (n+r) = 5n 2r + n = 5n 2r = 4n r = 2n Answer: 2n

Question 18

What is the number of positive integer solutions of x + y + z = 5?

Accepted Answer

The number of positive integer solutions to the equation x_1 + x_2 + + x_r = n is given by ^ n-1 C_ r-1 . For the given equation x + y + z = 5, we have n = 5 and r = 3. The number of positive integer solutions is ^ 5-1 C_ 3-1 = ^ 4 C_ 2 . ^ 4 C_ 2 = 4 3 2 1 = 6. Answer: 6

Permutation Combination — Previous Year Questions & Solutions