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
2621
Evaluate 50C48.
Answer:
1225
Step 1: Use the property nCr = nC(n-r) to simplify. 50C48 = 50C2. Step 2: Expand 50C2 = (50 × 49) / (2 × 1). Step 3: 25 × 49 = 1225.
2622
From a squad of 12 players, a playing 11 must be chosen. How many ways can this be done?
Answer:
12
Step 1: We need to choose 11 out of 12 players, which is 12C11. Step 2: Use the property nCr = nC(n-r). So, 12C11 = 12C1. Step 3: 12C1 = 12 ways. (Equivalently, you just choose 1 player to leave out).
2623
A shop has 8 different flavors of ice cream. In how many ways can a customer choose 3 distinct flavors?
Answer:
56
Step 1: Order of selection doesn't matter, so use combinations. Step 2: Find 8C3. Step 3: 8C3 = (8 × 7 × 6) / (3 × 2 × 1) = 56 ways.
2624
In how many ways can a team of 2 members be selected from 6 people?
Answer:
15
Step 1: Selecting a team does not involve order, so use combinations. Step 2: We need to choose 2 from 6, which is 6C2. Step 3: 6C2 = (6 × 5) / (2 × 1) = 15 ways.
2625
If 10Cx = 10C3, what is a possible value of x other than 3?
Answer:
7
Step 1: By the property of combinations, nCx = nCy implies either x = y or x + y = n. Step 2: Here, n = 10 and y = 3. Step 3: Since x is not 3, x + 3 = 10, which means x = 7.
2626
What is the value of nCn?
Answer:
1
Step 1: Use the combination formula nCr = n! / (r!(n - r)!). Step 2: Substitute r = n: n! / (n!(n - n)!) = n! / (n! × 0!). Step 3: Since 0! = 1, it simplifies to n! / n! = 1.
2627
Calculate the value of 10C8.
Answer:
45
Step 1: Use the property nCr = nC(n - r). Step 2: 10C8 = 10C(10 - 8) = 10C2. Step 3: 10C2 = (10 × 9) / (2 × 1) = 90 / 2 = 45.
2628
Evaluate 5C3 (Combinations of 5 objects taken 3 at a time).
Answer:
10
Step 1: The formula for nCr is n! / (r!(n - r)!). Step 2: Substitute n = 5 and r = 3: 5! / (3! × 2!). Step 3: (5 × 4 × 3!) / (3! × 2 × 1) = 20 / 2 = 10.
2629
In how many ways can 6 people be seated on 3 chairs?
Answer:
120
Step 1: Seating people implies arrangement, so use permutations. Step 2: We need to arrange 3 people out of 6, which is 6P3. Step 3: 6P3 = 6 × 5 × 4 = 120 ways.
2630
What is the relation between nPr and nCr?
Answer:
nPr = nCr × r!
Step 1: nCr = n! / (r!(n - r)!). Step 2: nPr = n! / (n - r)!. Step 3: Multiply nCr by r!: [n! / (r!(n - r)!)] × r! = n! / (n - r)! = nPr. So, nPr = nCr × r!.