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Quiz: Permutations and Combinations

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Number of Questions: 40

Question: 36 -

In how many ways can a cricketer can score 200 runs with fours and sixes only?

Options:

Question: 37 -

Ten participants are participating in a competition. In how many ways can the first three prizes be won?

Options:

920

680

720

820

Question: 38 -

Seven different objects must be divided among three people. In how many ways can this be done if one or two of them must get no objects?

Options:

381

180

Question: 39 -

A box contains 2 red balls, 3 black balls, and 4 white balls. Find the number of ways by which 3 balls can be drawn from the box in which at least 1 black ball should be present.

Options:

Question: 40 -

How many words, with or without meaning, can be formed using all letters of the word EQUATION using each letter exactly once?

Options:

40,320

38,400

39,320

38,320

Quiz: Permutations and Combinations

Solution:
200 runs can be scored by scoring only fours or through a combination of fours and sixes.
Possibilities are 50 x 4, 47 x 4 + 2 x 6, 44 x 4 + 4 x 6 ... A total of 17 ways

Solution:
Out of 10 participants, the first three prizes can be won in,
¹⁰P₃ = 10! / (10 - 3)!
= 10 x 9 x 8
= 720 ways

Solution:
The possible combination could be (1 black ball and 2 non-black balls), (2 black balls and 1 non- black ball), and (only 3 black balls).

Therefore the required number of combinations = (³C₁ * ⁶C₂) + (³C₂ * ⁶C₁) + (³C₃)
r, + + = 45+18+1 = 64

Solution:
The Word EQUATION has exactly 8 letters which are all different.
Therefore the number of words that can be formed = Number of permutations of 8 letters taken all at a time.
= P (8, 8) = 8!
= 8 × 7 x 6 × 5 x 4 x 3 x 2 × 1
= 40,320

Quiz: Permutations and Combinations

Solution: 200 runs can be scored by scoring only fours or through a combination of fours and sixes.Possibilities are 50 x 4, 47 x 4 + 2 x 6, 44 x 4 + 4 x 6 ... A total of 17 ways

Solution: Out of 10 participants, the first three prizes can be won in,10P3 = 10! / (10 - 3)!= 10 x 9 x 8= 720 ways

Solution: The possible combination could be (1 black ball and 2 non-black balls), (2 black balls and 1 non- black ball), and (only 3 black balls). Therefore the required number of combinations = (3C1 * 6C2) + (3C2 * 6C1) + (3C3)r, + + = 45+18+1 = 64

Solution: The Word EQUATION has exactly 8 letters which are all different.Therefore the number of words that can be formed = Number of permutations of 8 letters taken all at a time.= P (8, 8) = 8!= 8 × 7 x 6 × 5 x 4 x 3 x 2 × 1= 40,320

Solution:
200 runs can be scored by scoring only fours or through a combination of fours and sixes.
Possibilities are 50 x 4, 47 x 4 + 2 x 6, 44 x 4 + 4 x 6 ... A total of 17 ways

Solution:
Out of 10 participants, the first three prizes can be won in,
¹⁰P₃ = 10! / (10 - 3)!
= 10 x 9 x 8
= 720 ways

Solution:
The possible combination could be (1 black ball and 2 non-black balls), (2 black balls and 1 non- black ball), and (only 3 black balls).

Therefore the required number of combinations = (³C₁ * ⁶C₂) + (³C₂ * ⁶C₁) + (³C₃)
r, + + = 45+18+1 = 64

Solution:
The Word EQUATION has exactly 8 letters which are all different.
Therefore the number of words that can be formed = Number of permutations of 8 letters taken all at a time.
= P (8, 8) = 8!
= 8 × 7 x 6 × 5 x 4 x 3 x 2 × 1
= 40,320