Probability, Permutation and Combination

Question: The number of distinct permutations of the letters of the word "AMERICA" is how many times that of the word "CANADA"?

Solution:
In the word AMERICA there are 7 letters in total,
with A appearing 2 times and all other letters appearing 1 time each.

∴ Number of distinct permutations = 7!/2!
= (7 × 6 × 5 × 4 × 3 × 2 × 1)/2
= 7 × 6 × 5 × 4 × 3
= 2520

Again,
In the word CANADA there are 6 letters in total,
with A appearing 3 times and all other letters appearing 1 time each.

∴ Number of distinct permutations = 6!/3!
= (6 × 5 × 4 × 3 × 2 × 1)/(3 × 2 × 1)
= 6 × 5 × 4
= 120

Therefore, the number of arrangements of AMERICA is 2520/120 = 21 times the number of arrangements of CANADA.

So the number of distinct permutations of 'AMERICA' is 21 times the number of distinct permutations of 'CANADA'.

প্রশ্ন: 7 জন ব্যক্তিকে 1 টি গোলটেবিলের চারপাশে কতভাবে বসানো যাবে?

সমাধান: 
7 জন ব্যক্তিকে 1 টি গোলটেবিলের চারপাশে সাজানো যাবে (n - 1)! উপায়ে।
= (7 - 1)! 
= 6!
= 720

Question: A fair coin is flipped three times. What is the probability that the coin lands head each time?

Solution:
All possible outcome = {HHH, HHT, HTT, ΗΤΗ, ΤΗΗ, TTH, THT, TTT} = 8

It will be head every time, this occurs 1 time = {HHH}

∴ Probability = 1/8

Question: A committee of 4 members is to be selected from 6 men and 5 women. In how many ways can this be done if exactly 2 men must be selected?

Solution:
Here, to form the committee of 4 members, we need to select 2 men 
and (4 - 2) = 2 women.

Number of ways to select 2 men from 6 men:
⁶C₂ = 6!/[2! (6 - 2)!]
= {6 × 5 × 4!}/[2! × (4)!]
= (6 × 5)/(2 × 1)
= 3 × 5
= 15 ways

Number of ways to select 2 women from 5 women:
⁵C₂ = 5!/[2! (5 - 2)!]
= 5!/[2! × (3)!]
= {5 × 4 × 3!}/[2! × (3)!]
= (5 × 4)/(2 × 1)
= 5 × 2
= 10 ways

So, total number of ways = 15 × 10 = 150.

Question: In how many ways can a garland be made using 6 different flowers?

Solution:
We know,
For a circular garland, the formula for arrangements is (n - 1)!/2
Here, n = 6

∴ The number of ways to make a garland with 6 flowers = (6 - 1)!/2 
 = 5!/2 
= (5 × 4 × 3 × 2)/2 
= 60 

Here the order of choosing the elements doesn’t matter and this is a problem in combinations.
We have to find the number of ways of choosing 4 elements of this set which has 11 elements.
This can be done in ¹¹C₄ ways = 330 ways

In two throws of a dice, n(S) = (6 x 6) = 36
Let E = event of getting a sum
= {(3, 6), (4, 5), (5, 4), (6, 3)}
∴P(E)=n(E)/n(S)
=4/36
=1/9

Question: A volunteer group consists of 4 male volunteers and 6 female volunteers. If a team of 4 volunteers is to be chosen at random to lead the event, what is the probability that the team will include at least 2 women?

Solution:
Given,
Total people = 10

∴ Ways of selecting 4 people from 10 = ¹⁰C₄
= 210

We want at least 2 women, so there are 3 possible combinations:
1st combination: 2 women and 2 men
= ⁶C₂ × ⁴C₂
= 15 × 6
= 90

2nd combination: 3 women and 1 man
= ⁶C₃ × ⁴C₁
= 20 × 4
= 80

3rd combination: 4 women and 0 man
= ⁶C₄
= 15

∴ Total outcomes = 90 + 80 + 15
= 185

∴ Probability = 185/210
= 37/42

In a simultaneous throw of two dice, we have n(S) = (6 x 6) = 36

Then, E = {(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)}

∴ n(E) = 27
∴ P(E) = n(E)/n(S) = 27/36 = 3/4.

Question: How many ways the letters of the word 'TEACHER' can be arranged ?

Solution:
The word 'TEACHER' has 7 letters
Here, E = 2 times

We know, 
Number of distinct permutations = n!/(p1! × p2!......) 
= 7!/2!
= (7 × 6 × 5 × 4 × 3 × 2!)/2!
= 2520

∴ Distinct permutations 2520

∴ Probability = Possible outcomes/Total outcomes = 8/12 = 2/3

Hence, the required probability is 2/3.

Question: In a box, there are 7 yellow, 8 black, and 5 white balls. One ball is picked randomly. What is the probability that it is neither yellow nor white?

Solution:
মোট বলের সংখ্যা = 7 + 8 + 5 = 20 টি।

ধরি, E হলো এমন ঘটনা যেখানে বলটি হলুদ বা সাদা কোনোটিই নয়, অর্থাৎ বলটি কালো।
∴ অনুকূল ফলাফলের সংখ্যা, n(E) = 8

সম্ভাব্যতা = (অনুকূল ফলাফলের সংখ্যা)/(মোট ফলাফলের সংখ্যা)
= 8/20
= 2/5

অতএব, বলটি হলুদ বা সাদা না হওয়ার সম্ভাব্যতা হলো 2/5।

Required number of ways = ⁶C₁× ⁸C₃ + ⁶C₂× ⁸C₂ + ⁶C₃× ⁸C₁+ ⁶C₄× ⁸C₀
= {6 × (8 × 7 × 6)/3!} + {(6 × 5)/(2 × 1) × (8 × 7)/(2 × 1)} + {(6 × 5 × 4)/3! × 8} + {(6 × 5)/(2 × 1) × 1}
= (336 + 420 + 160 + 15)
= 931.

Probability of getting head on all tosses = 1/2 × 1/2 × 1/2 × 1/2 × 1/2 = 1/32

Let S be the sample space.
Total number of students in the class=12 boys + 16 girls = 28
Then, n(S) = 28
Let E be the event of calling one of them by enrollment number.
Given that, the number of girls = 16.
Then, n(E) = 16.
The probability that the one called is a girl = n(S)/n(E) = 16/28 = 4/7.

Question: In a box, there are 7 red, 8 blue, and 5 green balls. One ball is picked randomly. What is the probability that it is neither red nor green?

Solution:
মোট বলের সংখ্যা = 7 + 8 + 5 = 20 টি।

ধরি, E হলো এমন ঘটনা যেখানে বলটি লাল বা সবুজ কোনোটিই নয়, অর্থাৎ বলটি নীল।
∴ অনুকূল ফলাফলের সংখ্যা, n(E) = 8

সম্ভাব্যতা = (অনুকূল ফলাফলের সংখ্যা)/(মোট ফলাফলের সংখ্যা) = 8/20
= 2/5

অতএব, বলটি লাল বা সবুজ না হওয়ার সম্ভাব্যতা হলো 2/5

Question: A bag contains 7 green balls, 8 blue balls, and 5 yellow balls. One ball is drawn at random. What is the probability that the ball drawn is neither green nor yellow?
​
​Solution:
​Total balls = 7 + 8 + 5 = 20
​
​Favorable outcomes = balls that are neither green nor yellow, that mean blue balls = 8
​
​∴ P(blue) = Favorable outcomes​/total outcomes​ = 8/20 = 2/5

Question: Two dice are thrown simultaneously. What is the probability that the total score is an even number?

Solution:
Clearly, n(S) = (6 × 6) = 36

Let E = Event that the sum is an even number.
E = {(1, 1), (1, 3), (1, 5),
     (2, 2), (2, 4), (2, 6),
     (3, 1), (3, 3), (3, 5),
     (4, 2), (4, 4), (4, 6),
     (5, 1), (5, 3), (5, 5),
     (6, 2), (6, 4), (6, 6)}

∴ n(E) = 18

∴ P(E) = n(E)/n(S)
= 18/36
= 1/2

Question: In how many different ways can the letters of the word 'MEETING' be arranged so that the vowels always come together?

Solution:
In the word 'MEETING' we treat the three vowels 'E, E' and 'I' as one letter. Thus, we have MTNG(EEI) this has 5 letters.
As we know that number of ways of arranging n letters out of which r are of same type = n!/r!

∴ Number of ways of arranging these letters = 5! = 120
And 3 vowels (E, E,I) can be arranged in 3!/2! = 3 ways

∴  Required number of ways = 120 × 3 = 360

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

Solution:
Total letters = 5 (L, E, A, R, N)
Vowels = E, A (2 vowels)
Consonants = L, R, N (3 consonants)

Total arrangements of all 5 letters = 5! = 5 × 4 × 3 × 2 × 1 = 120

Treating the vowels (EA) as one single letter,
Now we have: (EA), L, R, N → 4 units.
These 4 units can be arranged in = 4! = 24 ways.

The two vowels EA can be arranged in their place in 2! = 2 ways
∴ Number of arrangements where vowels are together = 24 × 2 = 48 ways

∴ Number of arrangements where vowels are never together
= Total - Together
= 120 - 48
= 72

Question: In how many different ways can be letters of the word 'TENNIS' be arranged?

Solution:
TENNIS whereas total 6 letters and N comes two times.
So, arrangements are = 6!/2! = 720/2 = 360 ways

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 XTR (EA),
i.e., 4 letters.
These letters can be arranged in = 4! = 4 × 3 × 2 × 1 = 24 ways
The letters EA may be arranged amongst themselves in 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

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 how many ways can 4 guests from a group of 7 guests be seated around a circular table?

Solution:
7 জন থেকে 4 জন নির্বাচন করার উপায়:
⁷C₄ = 7!/(4! × 3!)
= (7 × 6 × 5)/(3 × 2 × 1)
= 210/6 = 35

4 জন ব্যক্তিকে একটি গোলাকার টেবিলে সাজানোর উপায় = (4 -1)!
= 3!
= 6

∴ মোট উপায় = 35 × 6 = 210

Let S be the sample space.
Then n(S) = no of ways of drawing 2 balls out of (6 + 4) = ¹⁰C₂
= (10 × 9)/(2 × 1) = 45
Let E = event of getting both balls of same colour
Then,
n(E) = no of ways (2 balls out of six) or (2 balls out of 4)
= ⁶C₂ + ⁴C₂
= (6 × 5)/(2 × 1) + (4 × 3)/(2 × 1)
= 15 + 6
= 21
Therefore,
P(E) = n(E)/n(S)
= 21/45
= 7/15

Multiple of 15 from 20 to 100 is 30, 45, 60, 75, 90 = 5
∴ Probability = 5/81

Question: A bag contains 45 marbles, 15 of which are red. If one marble is picked at random, what is the probability that it is not red?

Solution:
Given that, 
Total marbles = 45  
Red marbles = 15  
∴ Non-red marbles = 45 - 15 = 30

∴ Probability of picking a non-red marble = Number of non-red marbles/Total marbles  
= 30/45  
= 2/3

Question: In how many ways can 5 people from a group of 8 people be seated around a circular table?

Solution:
প্রথমে,
৮ জনের মধ্যে থেকে ৫ জন বেছে নেওয়ার উপায় = ⁸C₅
= 8!/{5!(8 - 5)!}
= 8!/(5! × 3!)
= 56

এবং,
এই ৫ জনকে বৃত্তাকার টেবিলে বসানোর উপায় = (5 - 1)!
= 4!
= 24

∴ মোট উপায় = 56 × 24
= 1344

Question: In how many ways can all the letters of the word 'LEADER' be arranged?

Solution:
এখানে, মোট বর্ণ সংখ্যা 6 টি যার মধ্যে 2 টি E এবং বাকি L, A, D, R প্রত্যেকটি 1 টি করে আছে।
∴ বিন্যাস সংখ্যা = 6!/2!
= (6 × 5 × 4 × 3 × 2!)/2!
= 6 × 5 × 4 × 3
= 360

Question: In a box, there are 7 yellow, 8 black, and 5 white balls. One ball is picked randomly. What is the probability that it is neither yellow nor white?

Solution:
মোট বলের সংখ্যা = 7 + 8 + 5 = 20 টি।

ধরি, E হলো এমন ঘটনা যেখানে বলটি হলুদ বা সাদা কোনোটিই নয়, অর্থাৎ বলটি কালো।
∴ অনুকূল ফলাফলের সংখ্যা, n(E) = 8

সম্ভাব্যতা = (অনুকূল ফলাফলের সংখ্যা)/(মোট ফলাফলের সংখ্যা)
= 8/20
= 2/5

অতএব, বলটি হলুদ বা সাদা না হওয়ার সম্ভাব্যতা হলো 2/5।

Total number of balls = (6 + 4 + 2 + 3)
= 15.
Let,
E be the event of drawing 2 red balls.
Then,
n(E) = ⁶C₂
= (6 × 5)/(2 × 1)
= 15.
Also, n(S) = ¹⁵C₂
= (15 × 14)/(2 × 1)
= 105.
∴ P(E) = n(E)/n(S)
= 15/105
= 1/7.

Question: Three unbiased coins are tossed. What is the probability of getting exactly two heads?

Solution:
তিনটি মুদ্রার জন্য মোট ফলাফল = 2³ = 8

মোট নমুনা বিন্দু = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
= 8 টি

ঠিক দুইটি হেড আসার অনুকূল ফলাফল = (HHT, HTH, THH)
∴ অনুকূল ঘটনার সংখ্যা = 3

∴ সম্ভাবনা = অনুকূল ঘটনা/মোট সম্ভাব্য ঘটনা
= 3/8

Question: There are 5 doors to a lecture room. Two are red, and the others are green. In how many ways can a lecturer enter the room and leave the room from different colored doors?

Solution:
There are 2 red and 3 green doors. We have two cases:
The room can be entered from a red door (2 red doors, so 2 ways) and can be left from a green door (3 green doors, so 3 ways): 2 × 3 = 6

The room can be entered from a green door (3 green doors, so 3 ways) and can be left from a red door (2 red doors, so 2 ways): 3 × 2 = 6 

Hence, the total number of ways = 6 + 6 = 12

Question: Tickets are numbered from 1 to 30. One ticket is drawn at random. What is the probability that the number is a multiple of 4 or 6?

Solution:
Total tickets = 30

Now, 
Multiples of 4 is-  4, 8, 12, 16, 20, 24, 28 = 7
multiples of 6 is- 6, 12, 18, 24, 30 = 5
And multiples of both (12, 24) = 2

∴ Favorable outcomes = 7 + 5 - 2 = 10

∴ Probability = favorable outcomes/total outcomes 
= 10/30
= 1/3

So the probability is 1/3.

The sample space when a dice is rolled, S = (1, 2, 3, 4, 5, and 6)
So, n (S) = 6
E is the event of getting an odd number.
So, n (E) = 3
Probability of getting an odd number P (E) = (Total number of favorable outcomes)/(Total number of outcomes)
n(E)/n(S) = 3/6
= 1/2

Total results = 2x + 3x + 4x = 9x,
Favourable result is (red ball) = 2x
∴ Probability = 2x/9x
= 2/9
Answer: 2/9

Let, rotten apples = 10 × 2/5 = 4, not rotten = 6

If 1 apple is rotten + 2 apples are not rotten 
= ⁴C₁ × ⁶C₂ = 60
If 2 apples are rotten + 1 apple is not rotten 
= ⁴C₂ × ⁶C₁ = 36
If 3 apples are rotten = ⁴C₃ = 4

Total outcomes = ¹⁰C₃ = 120
∴ Probability = (60 + 36 + 4) / 120 = 100/120 = 5/6