Here, S = {1, 2, 3, 4, 19, 20}.
Let E = event of getting a multiple of 3 or 5
= {3, 6 , 9, 12, 15, 18, 5, 10, 20}.
P(E) = n(E) / n(S)
= 9/20

Here, S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}
Let E = event of getting at most two heads.
Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}
P(E) = n(E)/n(S)
= 7/8.

Total number of balls = (8 + 7 + 6) = 21.
Let E = event that the ball drawn is neither red nor green
= event that the ball drawn is blue.
n(E) = 7
P(E) = n(E)/n(S)
= 7/21
= 1/3

Total number of outcomes possible when a die is rolled = 6 (∵ any one face out of the 6 faces).
Hence, Total number of outcomes possible when two dice are rolled, n(S) = 6 × 6 = 36.
Let E = the event of getting two numbers whose product is even.
= {(1,2), (1,4), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,2), (3,4), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,2),(5,4), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

Hence, n(E) = 27
P(E) = n(E) / n(S)
= 27/36
= 3/4

Total Number of letters in the word ASSASSINATION, n(S) = 13
Total number of consonants in the word ASSASSINATION = 7 (i.e., {S, S, S, S, N, T, N }
P(consonant) = 7/13

Total number of balls = 4 + 5 + 6 = 15
Let S be the sample space.
n(S) = Total number of ways of drawing 3 balls out of 15 = 15C3
Let E = Event of drawing 3 balls, all of them are yellow.
n(E) = Number of ways of drawing 3 balls, all of them are yellow = Number of ways of drawing 3 balls from the total 5 = 5C3 (∵ there are 5 yellow balls in the total balls)
P(E) = n(E)/n(S) = 2/91

Total number of outcomes possible when a die is rolled = 6 (∵ any one face out of the 6 faces).
Hence, Total number of outcomes possible when two dice are rolled, n(S) = 6 × 6 = 36.
Prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 … etc
Let E = the event that the total is a prime number
= {(1,1), (1,2), (1,4), (1,6), (2,1), (2,3), (2,5), (3,2), (3,4), (4,1), (4,3), (5,2), (5,6), (6,1), (6,5)}
Hence, n(E) = 15
P(E) = n(E)/n(S)
= 15/36 = 5/12

n(S) = Total number of ways of drawing 2 cards from 52 cards = 52C2.
Let E = event of getting 1 club and 1 diamond.
We know that there are 13 clubs and 13 diamonds in the total 52 cards.
Hence, n(E) = Number of ways of drawing one club from 13 and one diamond from 13 = ¹³C₁ × ¹³C₁ = 13/102
P(E) = n(E)/n(S)
= (¹³C₁ × ¹³C₁) / ⁵²C₂
= 13/102.

We may have ( 1 black and 2 non-black ) or ( 2 black and 1 non- black ) or ( 3 black ).
Required number of ways = (³C₁ x ⁶C₂) + (³C₂ x ⁶C₁) + (³C₃) = 64.

In the word ‘MATHEMATICS’, we treat the vowels AEAI as one letter. Thus, we have MTHMTCS (AEAI).
Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different.
Number of ways of arranging these letters = 8! / (2!)(2!) = 10080.
Now, AEAI has 4 letters in which A occurs 2 times and the rest are different. Number of ways of arranging these letters = 4! / 2! = 12.
Required number of words = (10080 x 12) = 120960.

1 red ball can be selected in ⁴C₁ ways 1 white ball can be selected in ³C₁ ways 1 blue ball can be selected in ²C₁ ways
Total number of ways = ⁴C₁ x ³C₁ x ²C₁ = 4 x 3 x 2 = 24.

• Please note that we have multiplied the combination results, we use to add when their is OR condition, and we use to multiply when there is AND condition.
• In this question it is AND as 1 red AND 1 White AND 1 Blue, so we multiplied

We need to find the ways that vowels NEVER come together.
Vowels are A, E Let the word be FTR(AE) having 4 words.
Total ways = 4! = 24
Vowels can have total ways 2! = 2
Number of ways having vowel together = 48
Total number of words using all letter = 5! = 120.
Number of words having vowels never together = 120 – 48 = 72

nPr = n! / (n−r)!
59P3 = 59! / (56)!
= 59 ∗ 58 ∗ 57 ∗ 56! / (56)!
=195054

Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.
The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.
The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.
Required number of ways = (1 x 5 x 4) = 20.

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4) = (7C₃ x 4C₂)
= { (7 x 6 x 5) / (3 x 2 x 1) } x { (4 x 3) / (2 x 1) } = 210.
Number of groups, each having 3 consonants and 2 vowels = 210.

Each group contains 5 letters.
Number of ways of arranging 5 letters among themselves = 5! = 5 x 4 x 3 x 2 x 1 = 120.

Required number of ways = (210 x 120) = 25200.