Probability, Permutation and Combination

Question: In how many ways can the letters of the word 'MISSISSIPPI' be arranged such that the first letter is always 'M'?

Solution:
The word 'MISSISSIPPI' contains 11 letters.

Condition: The first letter is fixed as 'M'.

So, the remaining 11 - 1 = 10 positions are to be filled by the remaining letters.

Among the remaining 10 letters, the repeated letters are:
I (4 times), S (4 times), P (2 times).

∴ Number of arrangements of the remaining 10 letters
= 10!/(4! × 4! × 2!)
= 3,628,800/(24 × 24 × 2)
= 3,628,800/1,152
= 3,150

Question: A bag contains 6 red, 5 blue, and 4 green balls. Two balls are drawn one after another without replacement. What is the probability that both balls are red?

Solution:
Total balls = 6 + 5 + 4 = 15
Probability of first red = 6/15
Probability of second red = 5/14

∴ Probability (both red) = (6/15) × (5/14)
= 30/210
= 1/7

Therefore, the probability that both balls are red is 1/7.

Question: If ⁷P_r = 210, what is the value of r?

Solution:
We know,
ⁿP_r = n!/(n - r)!

Given that,
⁷P_r = 210
⇒ 7!/(7 - r)! = 210
⇒ 5040/(7 - r)! = 210 (7! = 5040)
⇒ (7 - r)! = 5040/210
⇒ (7 - r)! = 24
⇒ (7 - r)! = 4!   ;(4! = 4 × 3 × 2 × 1 = 24)
⇒ 7 - r = 4
⇒ r = 7 - 4
∴ r = 3

10 থেকে 100 পর্যন্ত 5 এর গুণিতক সংখ্যাগুলো হলোঃ
10, 15, 20, 25, 30 ,35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95 এবং 100 = 19টি ।

এখন, 10 থেকে 100 পর্যন্ত 9 এর গুণিতক সংখ্যাগুলো হলোঃ
18, 27, 36, 45, 54, 63, 72, 81, 90 এবং 99 = 10 টি

10 থেকে 100 পর্যন্ত মোট সংখ্যা আছে = 100 - 10 + 1 = 91টি ।

এর মধ্যে, 45 এবং 90 দুইটা দুই জায়গাতেই আছে তাই একবার কাউন্ট হবে।

∴ নির্ণেয় সম্ভাব্যতা = (19+10-2)/91
= 27/91

Probability of getting a tail when a single coin is tossed 
=1/2
Probability of getting 4 when a die is thrown 
=1/6
Required probability,
=1/2 × 1/6
=1/12

Question: Find the number of triangles which can be formed by joining the angular points of a polygon of 8 sides as vertices.

Solution:
A triangle needs 3 points.
And a polygon of 8 sides has 8 angular points.

Hence,
Number of triangles formed
= ⁸C₃ 
= (8 × 7 × 6)/(3 × 2 × 1)
= 56

∴ 56 triangles can be formed by joining the angular points of a polygon of 8 sides as vertices

Question: All distinct permutations of the letters of the word PRIDE are arranged in alphabetical order. What is the rank of the word PRIDE?

Solution:
Here,
The order of each letter in the dictionary is DEIPR.

Now,
with D in the beginning, the remaining letters can be permuted = 4! ways.
= 24 ways

Similarly,
with E in the beginning, the remaining letters can be permuted = 4! ways.
= 24 ways

Similarly,
with I in the beginning, the remaining letters can be permuted = 4! ways.
= 24 ways

Now,
with P in the beginning and R in the second position,
the remaining letters are D, E, I.

Before R, there are D, E, I.

So permutations = 3 × 3!
= 3 × 6
= 18

Now,
with P and R fixed, the remaining letters are D, E, I.

Before I, there are D, E.

So permutations = 2 × 2!
= 2 × 2
= 4

Finally,
add the word itself.

Hence, the rank of the word PRIDE
= 24 + 24 + 24 + 18 + 4 + 1
= 95

14 জনের দল থেকে 1 জনকে ঠিক রেখে বাকি 13 জন থেকে (11 - 1) = 10 জনের টিম ¹³c₁₀ রুপে গঠন করা যাবে
= (13 × 12 × 11 × 10!)/(13 - 10)! × 10!
= (13 × 12 × 11)/3!
= (13 × 12 × 11)/(3 × 2 × 1)
= 13 × 2 × 11)
= 286
Answer: 286 উপায়ে টিম গঠন করা যাবে ।

Question: Mr. Shamim wants to arrange three out of his four saplings in a row on a shelf. If each sapling is in a pot of a different color, in how many different ways can he arrange the saplings?

Solution:
এখানে ভিন্ন ভিন্ন রঙ্গের পাত্রে থাকে, তার মানে নির্দিষ্ট রং ধারণ করে, তাই বিন্যাস হবে।
4 টি চারাগাছ হতে 3 টি নিয়ে সাজানো যায় = ⁴P₃ উপায়ে
= 4!/(4 - 3)! 
= 4! 
= 4 × 3 × 2 × 1 
= 24 উপায়ে

Question: There are 6 red, 5 blue, and 7 yellow balls in a bag. If a ball is picked at random, what is the probability of having either a red or a blue ball?

Solution:
ব্যাগে মোট বলের সংখ্যা = 6 + 5 + 7 = 18 টি।

লাল বল পাওয়ার সম্ভাবনা = 6/18
নীল বল পাওয়ার সম্ভাবনা = 5/18

∴ লাল অথবা নীল বল পাওয়ার সম্ভাবনা = (6/18) + (5/18)
​= (6 + 5)/18
​= 11/18

Question: In a social gathering, 12 people shake hands with each other. How many handshakes were there in total?

Solution:
একটি হ্যান্ডশেক সম্পন্ন করতে 2 জন ব্যক্তির প্রয়োজন হয়। তাই 12 জন ব্যক্তি থেকে 2 জনকে কতভাবে বাছাই করা যায়, সেটিই হবে মোট হ্যান্ডশেক সংখ্যা।

এখানে মোট ব্যক্তি, n = 12
প্রতিবার হ্যান্ডশেক করতে লাগে, r = 2

সূত্রমতে, মোট হ্যান্ডশেক সংখ্যা = ⁿC₂
 = n(n - 1)/2
= 12(12 - 1)/2
= (12 × 11)/2
= 132/2
= 66

A dice has 6 faces.
So there are 6 possible outcomes
Dice are rolled once AND then again.
So total possibilities = 6 x 6 = 36

The sum should be 8 of the 2 throws.
So which combination of numbers from 1 to 6 will yield us a sum of 8?
They are - (2,6); (6,2); (3,5); (5,3); (4,4)

So there are a total of 5 possibilities where the addition is 8.
But only 1 possibility where the first throw of dice is 4.

So, the Probability for the first throw to be 4 and sum to be 8 = 1/36.

Question: In how many ways can 4 people from a group of 7 people be seated around a circular table?

Solution: 
4 people out of 7 = ⁷C₄
= 7!/4!(7 - 4)!
= 7!/(4! × 3!)
= 35

And 4 people around a circular table = (4 - 1)! = 3! = 6

∴ Total ways = 6 × 35 = 210

Question: A committee of 3 men and 2 women is to be formed from 5 men and 4 women. In how many ways can the committee be formed?

​Solution:
We have 5 men and 4 women.
We need to choose 3 men from 5 and 2 women from 4.

∴ Number of ways = ⁵C₃ × ⁴C₂
= {5!/3!(5 - 3)!} × {4!/2!(4 - 2)!}
= {(5 × 4)/2} × {(4 × 3)/2}
= 10 × 6
= 60 ways

Question:  A small company employs 3 men and 5 women. If a team of 4 employees is to be randomly selected to organize the company retreat, what is the probability that the team will have exactly 2 women?

Solution:
the probability that the team will have exactly 2 women is = (⁵C₂ × ³C₂)/⁸C₄
= 30/70
= 3/7

Question: How many ways can the word QUESTION be arranged with 2 letters each time?

Solution:
'QUESTION' শব্দটিতে মোট 8 টি ভিন্ন ভিন্ন বর্ণ আছে (Q, U, E, S, T, I, O, N)

8টি ভিন্ন বর্ণ থেকে প্রতিবার 2টি বর্ণ নিয়ে সাজাতে হবে।

n টি ভিন্ন বস্তু থেকে r টি বস্তু নিয়ে সাজানোর উপায় = ⁿP_r = n!/(n - r)!

এখানে n = 8, r = 2
∴ ⁸P₂ = 8!/(8 - 2)!
= 8!/6!
= (8 × 7 × 6!)/(6!)
= 8 × 7
= 56

∴ 'QUESTION' শব্দের বর্ণগুলো থেকে প্রতিবার 2টি বর্ণ নিয়ে মোট 56 উপায়ে সাজানো যায়।

Question: A fruit shop has 12 types of fruits. You don’t like Mango and Papaya. How many ways can you select 5 different fruits from the ones you like?

Solution:
Given that, 
Total fruits = 12
Fruits you don’t like = 2
∴ Fruits you can choose = 12 - 2 = 10
Number of fruits to choose = 5

∴ Number of ways = ¹⁰C₅ = 10!/5!(10 - 5)!
= (10 × 9 × 8 × 7 × 6 × 5!)/(5! × 5!)
= (10 × 9 × 8 × 7 × 6)/(5 × 4 × 3 × 2)
= 252

So, there are 252 ways to select 5 different fruits from the ones you like.

Question: A box contains 24 electric bulbs, out of which 6 are defective. Two bulbs are chosen at random from this box. The probability that at least one of them is defective is:

Solution:
Given that,
Total bulbs = 24
Defective bulbs = 6
Non-defective bulbs = 24 - 6 = 18
Two bulbs are chosen at random (without replacement)

Now,
P(both non-defective) = (18/24) × (17/23)
= 306/552
= 51/92

And,
∴ P(at least one defective)
= 1 - P(both non-defective)
= 1 - (51/92)
= (92 - 51)/92
= 41/92

Question: In how many different ways can the letters of the word 'EXTRA' be arranged so that the vowels are never together?

Solution:
Taking the vowels (EA) as one letter, the given word has
the letters X T R (EA) = 4 letters.
∴ These letters can be arranged in = 4! = 24 ways.
∴ The letters EA may be arranged amongst themselves in = 2! = 2 ways.

∴ Number of arrangements having vowels together = (24 × 2) = 48 ways.

∴ Total arrangements of all letters = 5! = (5 × 4 × 3 × 2 × 1) = 120.

∴ Number of arrangements not having vowels together = (120 - 48) = 72

So number of arrangements where vowels are never together = 72

n(S) = number of ways of choosing 4 persons out of 9
= ⁹C₄ = (9 × 8 × 7 × 6)/(4 × 3 × 2 × 1) = 126

n(E) = number of ways of choosing 2 children out of 4 and 2 persons out of (3 + 2) personal
n(E) = ⁴C₂ × ⁵C₂ = {(4 × 3)/(2 × 1) × (5 × 4)/(2 × 1)} = 60

∴P(E) = n(E)/n(S) = 60/126 = 10/21.

Question: If a coin is tossed once, what is the probability of getting a tail?

Solution: Here, the total outcome is 2 (Head and Tail) 
The favorable outcome is 1 (Tail)

Therefore, Probability = Favorable outcome/Total outcome
= 1/2
= 0.5

Question: A bag contains 4 green balls and 3 yellow balls. If two balls are drawn without replacement, what is the probability that both are yellow?

Solution:
Total balls = 4 green + 3 yellow = 7 balls.
Probability that the first ball is yellow = 3/7

After removing 1 yellow ball, we have:
Remaining yellow balls = 2
Total remaining balls = 6
So, the probability that the second ball is yellow = 2/6 = 1/3

∴ Total probability (both yellow) = 3/7 × 1/3 = 1/7

Question: A fair coin is tossed 3 times. What is the probability of getting exactly two tails?

Solution:
একটি মুদ্রা 3 বার ছুঁড়লে মোট সম্ভাব্য ফলাফল (Total outcomes) হলো 2³ = 8 টি।
মোট ফলাফল (S) = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
মোট ফলাফলের সংখ্যা, n(S) = 8

'ঠিক দুটি টেইল' (exactly two tails) এর অনুকূল ফলাফল, E = {HTT, THT, TTH}
∴ অনুকূল ফলাফলের সংখ্যা, n(E) = 3

∴ সম্ভাবনা (Probability) = অনুকূল ফলাফলের সংখ্যা/মোট ফলাফল
P(E) = n(E)/n(S)
P(E) = 3/8

অতএব, ঠিক দুটি টেইল পাওয়ার সম্ভাবনা হলো 3/8।

When at least 2 women are included.
The committee may consist of 3 women, 2 men : It can be done in ⁴C₃ × ⁶C₂ : ⁴C₃ × ⁶C₂ ways.
or, 4 women, 1 man : It can be done in ⁴C₄ × ⁶C₁ : ⁴C₄ × ⁶C₁ ways.
or, 2 women, 3 men : It can be done in ⁴C₂ × ⁶C₃ : ⁴C₂ × ⁶C₃ ways.
Total number of ways of forming the committees,
= ⁴C₂ × ⁶C₃ + ⁴C₃ × ⁶C₂ + ⁴C₄ × ⁶C₁
= 6 × 20 + 4 × 15 + 1 × 6
= 120 + 60 + 6
= 186

Question: In how many ways can 6 books be selected from 10 books such that 2 particular books are always excluded?

Solution:
Since 2 specific books must always be excluded, we are effectively choosing from the remaining books.
Remaining books = 10 - 2 = 8 books

Now, we need to select 6 books from these 8 books.
The number of ways to do so is given by the combination formula = ⁸C₆
= 8!/(6! × (8 - 6)!)
= 8!/(6! × 2!)
= (8 × 7 × 6!)/(6! × 2 × 1)
= (8 × 7)/2
= 56/2
= 28

So, the total number of ways 28. 

The word,'APPLE' contans 5 letters, 1A, 2P, 1L and 1E.
∴ Required number of ways = 5!/2!
= 60

Question: If you toss a coin twice, what is the probability that you will get heads the second time?

Solution:
When we toss a coin the sample outcomes are = {HH, HT, TH, TT}
Total number of outcome = 4

Heads in the second time {HH, TH} = 2

∴ Probability that I will get heads the second time = 2/4 = 1/2 = (1 × 100)/2 % = 50%

Choose 1 person for the single room & from the remaining choose 2 for the double room & from the remaining choose 4 people for the four person room,
Then, 7C1 x 6C2 x 4C4
= 7 x 15 x 1
= 105

ⁿP_r = n!/(n-r)!
⁶P₃ = 6!/(6-3)!
⁶P₃ = 6!/3!
⁶P₃ = 120

Question: In a party there are 5 couples. Out of them 5 people are chosen at random. Find the probability that there are at the least two couples?

Solution:
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 × 6)/252
= 60/252
= 5/21

Since, no order to the committee is mentioned, a combination instead of a permutation is used.
Let's sort out what we have and what we want.
Have: 5 women, 6 men.
Want: 3 women AND 1 man.
The word AND means multiply.
Woman and Men

⁵C₃ × ⁶C₁
= 5!/(2! × 3!) × 6!/(5! × 1!)
= (5 × 2) × 6
= 60.

Question: How many Permutations of the letters of the word APPLE are there?

Solution:
Here,
APPLE = 5 letters.
But two letters PP is of same kind.
So, required permutations,
= 5!/2!
= 120/2
= 60

Question: A fair die is rolled once. What is the probability of getting an even number?

Solution:
A standard fair die has 6 faces.
{1, 2, 3, 4, 5, 6}

∴ Total possible outcomes = 6

And even numbers on a die, {2, 4, 6}
∴ Number of favorable outcomes = 3

∴ Probability of getting an even number = (Number of favorable outcomes)/(Total number of possible outcomes)
= 3/6
= 1/2

So the probability is 1/2.

Question: How many diagonals can be drawn in an octagon?

Solution:
An octagon has n = 8 vertices and 8 sides.
Total number of lines formed by joining 2 vertices out of 8 is: ⁸C₂
∴ Total lines = ⁸C₂
= 8!/{2! × (8 - 2)!}
= (8 × 7)/2 
= 28

These 28 lines include both the sides and the diagonals of the octagon.

Number of diagonals = Total lines - Number of sides
= 28 - 8
= 20

∴ The number of diagonals in an octagon is 20.

When 4 fair coins are tossed simultaneously, the total number of outcomes is 2⁴ = 16
At least 3 heads imply that one can get either 3 heads or 4 heads.
One can get 3 heads in ⁴C₃ = 4 ways and can get 4 heads in ⁴C₄

∴ Total number of favorable outcomes = 4 + 1 = 5
∴ The required probability = 1/4.

Question: In how many ways can 8 examination papers be arranged so that the best and the worst papers never come together?

Solution:
No. of ways in which 8 papers can be arranged = 8! ways.
When the best and the worst papers come together, regarding the two as one paper, we have only 7 papers.
These 7 papers can be arranged in 7! ways.
And two papers can be arranged themselves in 2! ways.

No. of arrangements when best and worst papers do not come together
= 8! - 7! × 2!
= 7!(8 - 2)
= 6 × 7!
= 30,240

Question: Fifteen distinct points are randomly placed on the circumference of a circle. At most how many triangles can be formed using these points?
​
​Solution:
​Given that,
​Number of distinct points = 15
​Maximum number of triangles = ¹⁵C₃
​= 15!/3!(15 - 3)!
​= (15 × 14 × 13 × 12!)/(3 × 2 × 12!)
= 455
​

Question: At a conference, everyone shakes hands with everybody else. If there were 190 handshakes, how many people were at the conference?

Solution:
Let the number of people at the conference be x.
A handshake occurs between any two people, which can be expressed using combinations.

According to the question,
^xC₂ = 190
⇒ x!/(2!(x - 2)!) = 190
⇒ {x(x - 1)(x - 2)!}/{2 × 1 × (x - 2)!} = 190
⇒ x(x - 1)/2 = 190
⇒ x(x - 1) = 380
⇒ x² - x - 380 = 0
⇒ x² - 20x + 19x - 380 = 0
⇒ x(x - 20) + 19(x - 20) = 0
⇒ (x - 20)(x + 19) = 0

So x - 20 = 0 or x + 19 = 0
∴ x = 20 or x = - 19

Since the number of people cannot be negative, x = 20 

Therefore, there were 20 people at the conference.

Question: There are 5 men and 4 women. Two members are to be selected for a committee. In how many ways can two  people be selected?

Solution:
Total candidates = 5 + 4 = 9

Number of ways to select 2 people = ⁹C₂ 
= 9!/2!(9 - 2)!
= (9 × 8 × 7!)/(2  × 7!)
= 36