Probability, Permutation and Combination

Question: If x and y are positive integers satisfying x + y = 7, what is the probability that x < y?

Solution: 
Since both x and y must be positive integers,
total possible ways = (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) = 6 pairs.
Next, we identify the pairs where x < y,
(1, 6), (2, 5), (3, 4) ; There are 3 pairs satisfying x < y.

∴ Probability = Number of pairs where x < y/Total number of pairs = 3/6 = 1/2

So the probability that x < y is 1/2

A team of 6 members has to be selected from the 10 players.
This can be done in 10C6 or 210 ways.
Now, the captain can be selected from these 6 players in 6 ways.
Therefore, total ways the selection can be made is = 210 × 6
= 1260

Given that a single dice is rolled.
Sample space = {1, 2, 3, 4, 5, 6}
Let P(A) be the probability of getting an odd number, where A = {1, 3, 5}
Let P(B) be the probability of getting a number less than 4, where B = {1,2,3}
A ⋂ B ={1, 3}
So, P(A) = 3/6 = 1/2
P(B) = 3/6 = 1/2
P(A ⋂ B) = 2/6 = 1/3

Let P be the required probability of getting an odd number or a number less than 4
P = P(A ⋃ B) = P(A) + P(B) - P(A ⋂ B)
Or, P(A ⋃ B) = 1/2 + 1/2 - 1/3
Or, P(A ⋃ B) = 1 - 1/3
Or, P(A ⋃ B) = 2/3

Therefore, the probability of rolling an odd number or a number less than 4 is 2/3.

There are 52 cards, out of which there are 12 face (Jack, King and Queen)) cards.
∴ Probability = 12/52 = 3/13

Question: How many 8 letter words can be formed by rearranging the letters of the word TRENDING such that T and G occupy the first and last positions respectively?

Solution:
As T and G should occupy the first and last position, the first and last position can be filled in only one following way.
T _ _ _ _ _ _ G.

The remaining 6 positions can be filled in the remaining words (R, E, N, D, I, N) where "N" comes twice.

Total permutations of these 6 letters with one letter repeating = 6!/2! = 720/2 = 360 ways

We know,
Probability = what we want/Total
Or = add; AND = multiply

We want a grey card OR ENGLISH card
There are 30 - 17 = 13 grey cards
There are 4 + 5 = 9 ENGLISH cards
Total cards = 30
Also, 5 grey cards are ENGLISH cards.

So Probability = 13/30 + 9/30 - 5/30 = (13 + 9 - 5)/30
= 17/30 [This subtraction is needed a grey card gets counted twice - once in 13 grey cards and once again in 9 ENGLISH cards.]

Question: In how many ways can a group of 2 teachers and 5 students be formed from 5 teachers and 8 students?

Solution:
We have 5 teachers and 8 students.
We need to choose 2 teachers from 5 and 5 students from 8.

∴ Number of ways = ⁵C₂ × ⁸C₅
= {5!/(2!(5 - 2)!)} × {8!/(5!(8 - 5)!)}
= {5!/(2!×3!)} × {8!/(5!×3!)}
= {(5×4)/(2×1)} × {(8×7×6)/(3×2×1)}
= 10 × 56
= 560 ways

Question: How many distinct arrangements can be made using all the letters of the word "MAMMAL" such that no two M's appear together?

Solution:
MAMMAL has 6 letters, where M = 3 times, A = 2 times and L = 1 time.
First arrange the letters other than M.
Number of arrangements of A, A, L = 3!/2! = 3

Now place the three M's in the gaps of these letters.
For example: _ A _ A _ L _
Total gaps = 4

Number of ways to choose 3 gaps for M = ⁴C₃ = 4

∴ Required number of arrangements = 3 × 4
= 12

Question: How many permutations of nine different digits may be made?

Solution:
আমরা জানি,
n সংখ্যক ভিন্ন জিনিস বা অক্ষর থেকে সবগুলি নিয়ে বিন্যাস সংখ্যা (Permutation) হলো n!
এখানে মোট অক্ষর সংখ্যা, n = 9
∴ নির্ণেয় বিন্যাস সংখ্যা = 9!

Question: A jar contains 6 white balls, 4 red balls, and 2 black balls. Two balls are drawn one after the other without replacement. Find the probability that both balls drawn are white.

Solution:
A jar contains 6 white balls, 4 red balls and 2 black balls.
∴ Total balls = 6 + 4 + 2 = 12
Two balls are drawn successively without replacement.

Now, 
Probability first ball is white = 6/12 = 1/2
After drawing one white ball, 5 white balls remain and the total number of balls remaining = 11
Probability second ball is white = 5/11

Since the draws are dependent (without replacement), multiply the probabilities.
∴ P(both white) = (6/12) × (5/11)
= (1/2) × (5/11)
= 5/22

So the probability that both balls are white is 5/22.

Question: If ⁿC_r = 7 and ⁿP_r = 840, then r! =?

Solution: 
We know,
r! × ⁿC_r = ⁿP_r
⇒ r! × 7 = 840
⇒ r! = 120
⇒ r! = 5 × 4 × 3 × 2 × 1
∴ r! = 5!

The digits to be used are 0,6 and 9
The required numbers are from 1 to 99999
The numbers are five digit numbers.
Therefore, every place can be filled by 0, 6 and 9 in 3 ways.
Total number of ways = 3 × 3 × 3 × 3 × 3 = 3⁵
But 00000 is also a number formed and has to be excluded.
Total number of numbers,
= 3⁵ - 1
= 243 - 1
= 242

We are to choose 11 players including 1 wicket keeper and 4 bowlers or 1 wicket keeper and 5 bowlers.
Number of ways of selecting 1 wicket keeper, 4 bowlers and 6 other players in ²C₁ × ⁵C₄ × ⁹C₆ = 840
Number of ways of selecting 1 wicket keeper, 5 bowlers and 5 other players in ²C₁ × ⁵C₅ × ⁹C₅ = 252
Total number of ways of selecting the team = 840 + 252 = 1092

As the toys are distinct and not identical,
For each of the 8 toys, we have three choices as to which child will receive the toy.
Therefore, there are 5⁸ ways to distribute the toys.
Hence, it is 5⁸ and not 8⁵.

If the student answers 4 questions out of the first five questions he can choose 6 questions from the remaining 8 questions.
Number of combinations will be = ⁵C₄ × ⁸C₆ = 140
If the student answers 5 questions from the first five questions he can choose 5 questions from the remaining 8 questions; Number of combinations will be = ⁵C₅ × ⁸C₅ = 56
So, total number of choices are = 140+56 =196

Question: In a box, there are 9 green balls, 5 blue balls, and 4 yellow balls. One ball is picked at random. What is the probability that it is neither green nor yellow?

Solution: 
Here,
Green balls = 9
Blue balls = 5
Yellow balls = 4
Total balls = 9 + 5 + 4 = 18

Let, E = event that the ball drawn is neither green nor yellow
= event that the ball drawn is blue

অর্থাৎ বলটি যদি সবুজ বা হলুদ না হয় তাহলে বলটি হবে নীল।

∴ n(E) = 5 
∴ P(E) = 5/18

Question: From a pack of cards two cards are drawn one after the other, with replacement. The probability that the first is a red card and the second is a king is -

Solution: 
Total cards in a pack = 52
Red cards = 26 (13 hearts + 13 diamonds)
Kings = 4 (2 red, 2 black)
Red kings = 2

Now,
Probability of first card being red,
∴ P(red) = 26/52 = 1/2

And,
Probability of second card being a king (after replacement),
∴ P(king) = 4/52 = 1/13

∴ P(first red and second king) = P(first red) × P(second king) = (1/2) × (1/13) = 1/26

Question: In how many ways can we select a team of 5 students from a given choice of 20?

Solution:
The number of possible ways of selection is given by,
20C5
​= 20!/5!(20 - 5)!
= (20 × 19 × 18 × 17 × 16 × 15!)/(5 × 4 × 3 × 2 × 15!)
​= 15504

So, the number of ways to select 5 students from 20 is 15504.

Question: Two dice are rolled together. What is the probability that the sum is at least 10?

Solution:
 দুইটি ছক্কা নিক্ষেপ করা হলে নমুনা বিন্দুর সংখ্যা = 6² = 36

প্রশ্নমতে,
দুইটি ছক্কায় উঠা সংখ্যাদ্বয়ের যোগফল ≥ 10 
 
এখন,
যোগফল 10 হলে অনুকূল ঘটনা = (4, 6), (5, 5), (6, 4) অর্থাৎ 3 টি। 
যোগফল 11 হলে অনুকূল ঘটনা = (5, 6), (6, 5) অর্থাৎ 2 টি । 
যোগফল 12 হলে অনুকূল ঘটনা = (6, 6) অর্থাৎ 1 টি। 

∴ মোট অনুকূল ঘটনা = 3 + 2 + 1 = 6

সম্ভাবনা = 6/36
= 1/6

Question: For the word 'MAGIC' how many different types of arrangement are possible so that the vowels are always together?

Solution:
In the Word MAGIC 
There are 2 vowels: A, I 
They can be arranged in 2! = 2 ways

There are three consonants: M, G, C
As the vowels are always together, we consider them as 1 letter.
So, 4 letter can be arranged in 4! = 24 ways

∴ Total number of arrangement is 2 × 24 = 48 words

Question: Using the digits 2, 5, and 7 exactly once each, three-digit numbers are formed. One number is selected at random. What is the probability that it is divisible by 4?

Solution:
Total numbers = 3! = 6
Numbers are 257, 275, 527, 572, 725, 752

We know, 
A number is divisible by 4 if the number formed by its last two digits is divisible by 4.

∴ Check last two digits are 72 and 52 divisible by 4.

∴ Favourable outcomes = 572 and 752 = 2

∴ Probability = 2/6 = 1/3

Number of ways of (selecting at least two couples among five people selected) = (⁵C₂ × ⁶C₁)
As remaining person can be any one among three couples left.
Required probability = (⁵C₂ × ⁶C₁)/¹⁰C₅
= (10 x 6)/252
= 5/21

Question: Two dice are thrown simultaneously. Find the probability of getting a multiple of 2 on one dice and multiple of 3 on the other dice.

Solution: 
Let x be required events and S be the sample space
Total outcomes when two dice are thrown,
6 × 6 = 36 
n(S) = 36
And then x = {(2, 3), (2, 6), (4, 3), (4, 6), (6, 3), (6, 6), (3, 2), (6, 2), (3, 4), (6, 4), (3, 6)}
n(x) = 11

Hence, required probability
= n(x)/n(S) = 11/36

Question: Find the number of triangles that can be formed by joining the angular points of a polygon of 10 sides as vertices.

Solution: 
A triangle needs 3 points.
A polygon of 10 sides has 10 angular points.

Hence, the number of triangles formed = ¹⁰C₃
= (10 × 9 × 8)/(3 × 2 × 1)
= 3 × 5 × 8
= 120

We want 2 red and 1 blue shirt
There are 4 red shirts and 3 blue shirts
Total = 15 shirts
You can choose a blue shirt 1^st, then a red shirt and then a red shirt

Probability = 3/15 × 4/14 × 3/13 = 6/455

OR You can choose a red shirt 1^st, then redshirt and then a blue shirt
OR You can choose a red shirt 1^st, then a blue shirt and then red-shirt
For all 3 the probability remains the same = 6/455

We need to add these 3 probabilities to get the total probability
Total probability = 6/455 + 6/455 + 6/455 = 18/455.

Question: You have 4 favorite novels on your shelf. If you decide to arrange these 4 novels in every possible order and it takes 45 seconds to move one novel, how much time will it take to arrange all possible combinations?

Solution:
4 novels can be arranged in = 4! ways
= 4 × 3 × 2 × 1 = 24 ways

Time required to move one novel = 45 seconds
time to arrange 4 novels = 45 × 4 = 180 seconds

So, total time required = 24 × 180 seconds
= 4320 seconds
= 4320/60
= 72 minutes

∴ It will take 72 minutes to arrange all possible combinations.

Question: How many 3-digit numbers can be formed using the digits 3, 4, 5, 6 and 7 without repetition?

Solution:
যেহেতু, অঙ্কের সংখ্যা = 5টি
এদের থেকে 3 অঙ্কবিশিষ্ট সংখ্যা গঠন করতে হবে (প্রতিটি অঙ্ক একবারই ব্যবহার করা যাবে)
∴ মোট সংখ্যা = ⁵P₃
= 5!/(5 - 3)!
= 5!/2!
= (5 × 4 × 3 × 2 × 1)/(2 x 1)
= 5 × 4 × 3
= 60

Question: In a group of 150 people, 90 people read Newspaper A, 65 people read Newspaper B, and 30 people read both Newspaper A and Newspaper B. How many people read neither Newspaper A nor Newspaper B?

Solution:
মোট লোক = 150
A পত্রিকা পড়ে, n(A) = 90
B পত্রিকা পড়ে, n(B) = 65
উভয়টি পড়ে, n(A ∩ B) = 30

∴ কমপক্ষে একটি পত্রিকা পড়ে, n(A ∪ B) = n(A) + n(B) - n(A ∩ B)
= 90 + 65 - 30
= 125

∴ যারা কোনটিই পড়ে না = মোট লোক - যারা কমপক্ষে একটি পড়ে
= 150 - 125
= 25

∴ 25 জন কোন পত্রিকাই পড়ে না।