Mathematics

Permutation Questions

Chapter: Algebra | Total Available: 26 Real Questions

Question 1 of 10 Authentic UPSC NDA Question
How many odd numbers between 300 and
400 are there in which none of the digits is
repeated?
Step-by-Step Logic & Solution

Question 2 of 10 Authentic UPSC NDA Question

Let A = {1, 2, 3, 4, 5} and B = {6, 7}. What is the number of onto functions from A to B?

Step-by-Step Logic & Solution

n(A) = 5 and n(B) = 2

 Number of onto functions = 25 – 2C1(2 – 1)5 =

 32 – 2 = 30

Question 3 of 10 Authentic UPSC NDA Question

In how many ways can the letters of the word INDIA be permutated such that in each combination, vowels should occupy odd positions?

Step-by-Step Logic & Solution

Question 4 of 10 Authentic UPSC NDA Question

The letters of the word EQUATION are arranged in such a way that all vowels as well as consonants are together. How many such arrangements are there?

Step-by-Step Logic & Solution

Question 5 of 10 Authentic UPSC NDA Question

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?

Step-by-Step Logic & Solution

Question 6 of 10 Authentic UPSC NDA Question

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

Step-by-Step Logic & Solution

Explanation: Number of arrangements = 3!2! = 12

Question 7 of 10 Authentic UPSC NDA Question

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

Step-by-Step Logic & Solution

Question 8 of 10 Authentic UPSC NDA Question

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)?

Step-by-Step Logic & Solution

Explanation: Last two digit of number which is divisible by 4 are 12, 24, 32

No. of number whose last two digits is 12 = 2!

No. of number whose last two digits is 24 = 2!

No. of nubmer whose last two digits is 32 = 2! 

Total number = 2! + 2! + 2! = 6

Question 9 of 10 Authentic UPSC NDA Question

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

Step-by-Step Logic & Solution

Solution

Option (b) is correct.

Explanation:

To determine how many numbers greater than 1000 can be formed using the digits 0, 1, 2, and 3 without repeating any digits, follow these steps: - The first digit (thousands place) must be among {1, 2, 3} (3 choices). - The second digit can be any of the remaining 3 digits. - The third digit can be any of the remaining 2 digits. - The fourth digit can be the last remaining digit. Using the multiplication principle: \[ \text{Total Numbers} = (\text{Choices for the first digit}) \times (\text{Choices for the second digit}) \times (\text{Choices for the third digit}) \times (\text{Choices for the fourth digit}) \] This can be expressed as: \[ \text{Total Numbers} = 3 \times 3 \times 2 \times 1 \] Calculating this:
\[ \text{Total Numbers} = 3 \times 3 \times 2 \times 1 = 18 \]
Thus, the total numbers greater than 1000 formable are 18.
📌 Hints / Properties Used:
  • Counting principle (multiplication principle).
  • Permutations of digits without repetition.
📊 Visual Diagram Suggestion:

A flow diagram showing choices for first, second, third, and fourth digits with branching paths can illustrate the counting process clearly.

📖 Factual Verification & Reference:

Verified against NCERT Class IX Mathematics, Chapter on Permutations and Combinations.

Question 10 of 10 Authentic UPSC NDA Question

How many 5-digit prime numbers can be formed using the digits 1, 2, 3, 4, 5 if the repetition of digits is not allowed?

Step-by-Step Logic & Solution

Today's UPSC Current Affairs

Padma Bhushan Call Deepens India Election Commission Row - streamlinefeed.co.ke

Read Today's Digest