All Categories MCQs
Topic Notes: All Categories
General Description
Plato
- Biography: Ancient Greek philosopher (427–347 BCE), student of Socrates and teacher of Aristotle, founder of the Academy in Athens.
- Important Ideas:
- Theory of Forms
- Philosopher-King
- Ideal State
2591
How many 3-digit numbers can be formed using 0, 1, 2, 3, 4 if repetition is allowed?
Answer:
100
Step 1: A 3-digit number cannot begin with 0. So, the hundreds place has 4 options (1,2,3,4). Step 2: The tens and units places can each take any of the 5 digits (0,1,2,3,4). Step 3: Total ways = 4 × 5 × 5 = 100.
2592
How many 4-digit numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if repetition is allowed?
Answer:
1296
Step 1: There are 4 positions to fill. Step 2: Since repetition is allowed, any of the 6 digits can go into any position. Step 3: Total ways = 6 × 6 × 6 × 6 = 1296.
2593
How many 3-digit numbers can be formed using the digits 1, 2, 3, 4, 5 if repetition is allowed?
Answer:
125
Step 1: There are 3 spots to fill. Step 2: Each spot can be filled in 5 ways since repetition is allowed. Step 3: Total ways = 5 × 5 × 5 = 125.
2594
What is the total number of 5-digit numbers that can be formed from 1, 2, 3, 4, 5 without repetition?
Answer:
120
Step 1: We are arranging all 5 distinct digits. Step 2: This is calculated by 5!. Step 3: 5! = 5 × 4 × 3 × 2 × 1 = 120.
2595
How many 4-digit numbers divisible by 5 can be formed from 1, 2, 3, 4, 5 without repetition?
Answer:
24
Step 1: For a number to be divisible by 5, its last digit must be 0 or 5. Here, only 5 is available (1 option). Step 2: The remaining 3 spots are filled by the remaining 4 digits in 4P3 ways. Step 3: Total ways = 4P3 × 1 = 4 × 3 × 2 = 24.
2596
How many 4-digit numbers greater than 3000 can be formed using 1, 2, 3, 4, 5 without repetition?
Answer:
72
Step 1: To be >3000, the thousands digit must be 3, 4, or 5 (3 options). Step 2: The remaining 3 places can be filled by the remaining 4 digits in 4P3 ways. Step 3: Total ways = 3 × (4 × 3 × 2) = 3 × 24 = 72.
2597
How many 3-digit even numbers can be formed using 0, 1, 2, 3, 4, 5 without repetition?
Answer:
52
Step 1: Two cases based on the last digit. Case 1: Last digit is 0 (1 way). First two digits can be filled in 5P2 = 20 ways. Step 2: Case 2: Last digit is 2 or 4 (2 ways). Hundreds place cannot be 0, so it has 4 options. Tens place has 4 options. 4 × 4 × 2 = 32 ways. Step 3: Total ways = 20 + 32 = 52.
2598
How many 4-digit numbers can be formed using the digits 0, 1, 2, 3, 4 without repetition?
Answer:
96
Step 1: A 4-digit number cannot start with 0. So, the thousands place has 4 options (1,2,3,4). Step 2: The remaining 3 places can be filled by the remaining 4 digits (including 0) in 4P3 ways. Step 3: Total ways = 4 × (4 × 3 × 2) = 4 × 24 = 96.
2599
How many 3-digit odd numbers can be formed from the digits 1, 2, 3, 4, 5 without repetition?
Answer:
36
Step 1: The last digit must be odd. From 1,2,3,4,5, the odd digits are 1, 3, 5 (3 options). Step 2: The remaining 2 spots are filled from the remaining 4 digits (4P2 ways). Step 3: Total ways = 4P2 × 3 = (4 × 3) × 3 = 12 × 3 = 36.
2600
How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5 without repetition?
Answer:
24
Step 1: For a number to be even, its last digit must be even. From 1,2,3,4,5, the even digits are 2 and 4 (2 options). Step 2: After choosing the last digit, 4 digits remain to fill the first 2 spots (hundreds and tens). Step 3: Number of ways = 4P2 × 2 = (4 × 3) × 2 = 12 × 2 = 24.