Probability, Permutation and Combination

Question: ⁷P_r = 210 and ⁷C_r = 35 then what is the value of r?

Solution:
Given that,
⁷P_r = 210 and ⁷C_r = 35

We know that,
ⁿP_r = r! × ⁿC_r
⇒ 210 = r! × 35
⇒ r! = 210/35
⇒ r! = 6
⇒ r! = 3!
∴ r = 3

Question: Everyone present at a party shakes hands with each other. If the total number of handshakes is 66, how many people were present at the party?

Solution:
ধরি, পার্টিতে মোট x জন লোক উপস্থিত ছিল।

 যেহেতু করমর্দন করতে দুইজন লোকের প্রয়োজন হয়, তাই মোট করমর্দন সংখ্যা = ^xC₂
= x!/{2!(x - 2)!}
= {x(x - 1)(x - 2)!}/{2!(x - 2)!}
= x(x - 1)/2
= (x² - x)/2

 প্রশ্নমতে,
(x² - x)/2 = 66
⇒ x² - x - 132 = 0
⇒ (x - 12)(x + 11) = 0
∴ x = 12 or x = - 11

কিন্তু, x = - 11 হতে পারে না।
∴ x = 12

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: In how many different ways can a committee of 3 members be selected from 5 people if a particular person must always be included in the committee?

Solution: 
Since one person must always be in the committee, we need to select the other two members from the remaining 4 people.
∴ Number of ways to choose the other two members = ⁴C₂
= (4 × 3)/(1 × 2) 
= 6  

Question: In a bag, there are 5 red, 3 green, and 2 blue marbles. If two marbles are drawn one after the other without replacement, what is the probability that the first one is red and the second one is green?

Solution:
মোট মার্বেলের সংখ্যা = 5 (লাল) + 3 (সবুজ) + 2 (নীল) = 10টি
প্রথম মার্বেলটি লাল হওয়ার সম্ভাবনা = 5/10 = 1/2
প্রথম মার্বেলটি তোলার পর থলেতে মোট মার্বেলের সংখ্যা = 10 - 1 = 9টি
দ্বিতীয় মার্বেলটি সবুজ হওয়ার সম্ভাবনা = 3/9 = 1/3

∴ প্রথমটি লাল এবং দ্বিতীয়টি সবুজ হওয়ার সম্ভাবনা = (প্রথমটি লাল হওয়ার সম্ভাবনা) × (দ্বিতীয়টি সবুজ হওয়ার সম্ভাবনা)
= 1/2 × 1/3
= 1/6

Question: What is the probability of getting a sum 7 from two throws of a dice?

Solution:
দুটি ছক্কা নিক্ষেপ করলে মোট সম্ভাব্য ফলাফল (Sample Space), n(S) = 6 × 6 = 36

ধরি, E হলো প্রাপ্ত সংখ্যাদ্বয়ের যোগফল 7 হওয়ার ঘটনা।
∴ অনুকূল ফলাফলগুলো হলো, E = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}

এখানে, n(E) = 6

আমরা জানি,সম্ভাবনা P(E) = n(E)/n(S)
= 6/36
= 1/6

Question: In how many ways can 5 examination papers be arranged so that the best and the worst papers never come together?

Solution:
The number of ways in which 5 papers can be arranged is 5! ways.
When the best and the worst papers come together, regarding the two as one paper, we have only 4 papers.
These 4 papers can be arranged in 4! ways.
And two papers can be arranged themselves in 2! ways.

∴ Number of arrangements when the best and worst papers do not come together,
= 5! - (4! × 2!)
= (5 × 4 × 3 × 2 × 1) - (4 × 3 × 2 × 1 × 2)
= 120 - 48
= 72

Question: Two fair dice are thrown together. What is the probability that the product of the two numbers that appear is 20?

Solution:
Total number of possible outcomes when throwing two dice = 6 × 6 = 36
Favorable outcomes where the product is 20.
The possible pairs (first die, second die) are, (4, 5) and (5, 4)
∴ favorable outcomes = 2
∴ Probability = Number of favorable outcomes/Total outcomes
= 2/36
= 1/18

∴ The probability is 1/18.

Question: In a simultaneous throw of a pair of dice, what is the probability of getting a total more than 9?

Solution:
When two fair six-sided dice are thrown together. Then we get,
Total outcomes = 6 × 6 = 36

And, Count outcomes with sum > 9

We want sum > 9 ⇒ sums = 10, 11, 12
Sum 10 = (4, 6), (5, 5), (6, 4) ⇒ 3 outcomes
Sum 11 = (5, 6), (6, 5) ⇒ 2 outcomes
Sum 12 = (6, 6) ⇒ 1 outcome

∴ Total favorable outcomes = 3 + 2 + 1 = 6

∴ Probability(sum > 9) = favorable outcomes/total outcomes
= 6/36
= 1/6

So the probability of getting a total more than 9 is 1/6.

Question: Two dice are tossed at once. What is the probability that their combined total is no less than 10?

solution:
When two fair six-sided dice are tossed, the total number of possible outcomes = 6 × 6 = 36.
We want the probability that the sum is ≥ 10 (i.e., 10, 11, or 12).

List all favorable outcomes,
Sum = 10: (4, 6), (5, 5), (6, 4) = 3 ways
Sum = 11: (5, 6), (6, 5) = 2 ways
And Sum = 12: (6,6) = 1 way

∴ Total favorable outcomes = 3 + 2 + 1 = 6

∴ Probability = favorable outcomes/total outcomes
= 6/36
= 1/6

So the probability is 1/6.

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

Solution:
5 people out of 8 = 8C5
= 8!/5!(8 - 5)!
= 8!/(3! × 5!)
​= (8 × 7 × 6 × 5!)/(6  × 5!)
= 56

And 5 people around a circular table = (5 - 1)! = 4! = 24

∴ Total ways = 24 × 56 = 1344

Question: Two unbiased coins are tossed. What is the probability of getting at most one head?

Solution:
Total cases = {HH, HT, TH, TT} = 4
Favorable cases = {TT, HT, TH} = 3

∴ Required Probability = 3/4

'DARKER' has 6 letters.

Thus, we can arrange 6 letters in 6! ways.
But R gets repeated. There are 2R's. So divide by 2!

∴ Total ways = 6!/2! = 360
Vowels not together = Total ways - Vowels together

Consider the 2 vowels (A and E) as one group.
We have 4 letters and 1 group = 5
We can arrange them in 5! Ways.

But again here R comes twice. So we will have 5!/2!
Also, the 2 vowels can be arranged in 2! Ways.
SO the number of ways with vowels together = 2! × (5!/2!) = 120

∴ Number of ways with vowels not together = 360 - 120 = 240.

Question: Two cards are drawn at random and without replacement from a standard deck of 52 cards. What is the probability that both cards are face cards?

Solution:
Total card = 52
Total face card = 3 × 4 = 12

Total ways to choose 2 cards from 52 = ⁵²C₂ = (52 × 51)/2 = 1326

Total ways to choose 2 face cards from 12 = ¹²C₂ = (12 × 11)/2 = 66

∴ So, the probability that both cards are face cards = 66/1326
= (2 × 3 × 11)/(2 × 3 × 221)
= 11/221

All possible outcome = {HHH, HHT, HTT, HTH, THH, TTH, THT, TTT} =  8
It will be head every time, this occurs 1 time
∴ Probability =  1/8

Question: Find the value of  0!/√4.

Solution:
According to the factorial law,
n! = n × (n - 1)!

If n = 1, then
1! = 1 × (1 - 1)!
⇒ 1! = 1 × 0!
⇒ 1 = 0!
∴ 0! = 1

Now,
0!/√4 = 1/√(2²)
= 1/2

যে কোন করমর্দন অথবা কোলাকুলির অংকে শুধু কত জন করমর্দন (Handshake), বা কোলাকুলি করল তা দেয়া থাকবে
এক্ষেত্রে মনে রাখতে হবে যে প্রত্যেক বার করমর্দন বা কোলাকুলি করার সময় মোট ২ জন লোকের প্রয়োজন।
তাই এক্ষেত্রে সূত্রটি হবে ⁿC₂=^{মোট লোক}C_{২ জন সব সময়}
20C₂= 20!/2!(20 - 2)!
= 20!/2!18!
= (20 × 19)/2
= 10 × 19
= 190

Choosing 4 marbles means 1^st marble AND then 2^nd AND then 3^rd AND then 4^th marble
At least 1 red marble means there can be 1, 2, 3, or 4 red marbles.
So, Probability of choosing red = 1 - Probability of not choosing red
If we remove red marbles, then 15 - 4 red marbles = 11 marbles remain.
We have to choose 4 out of these 11.

So Probability of choosing 4 marbles that are not red = 11/15 × 10/14 × 9/13 × 8/12 = 22/91.

∴ Probability of picking at least 1 red marble = 1 - 22/91 = 69/91.
[Here we reduce the denominator i.e. the total number of marbles because once we remove a marble from the box we do not put it back in the box.
So while removing the 2^nd marble, there are only 15 - 1 = 14 marbles in the box. The same is for the 3^rd and 4^th marbles.]

Question: A bag contains 5 green, 3 yellow, and 2 black balls. If one ball is drawn at random, what is the probability that it will not be a yellow ball?

Solution:
Given that,
Green balls = 5
Yellow balls = 3
Black balls = 2

∴ Total balls = 5 + 3 + 2 = 10
∴ Number of non-yellow balls = Green + Black = 5 + 2 = 7

We know,
Probability(not yellow) = favorable outcomes/total outcomes
= 7/10

∴ The probability that the ball drawn is not yellow is 7/10.

S = {1, 2, 3,............, 49, 50}
E = {4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 7, 14, 21, 35, 42, 49}
n(S) = 50
n(E) = 18

P(E) = n(E)/n(S)
= 18/50
= 9/25.

Here, the order of arrangement of digits does matter.
ⁿP_r = n!/(n-r)!
ⁿP_r = 4!/(4-3)!
⁴P₃ = 4!/1!
⁴P₃ = 4!
⁴P₃ = 24

Question: In how many ways can 5 books be selected from 12 books, with 2 particular books always left out?

Solution:
মোট বস্তু, n = 12
সর্বদা বাদ বা বর্জন থাকবে, m = 2
এবং প্রতিবার নিতে হবে, r = 5

∴ সমাবেশ = ^{(n - m)}C_r = ^{(12 - 2)}C₅
= ¹⁰C₅
= 10!/[5!(10 - 5)!]
= (10 × 9 × 8 × 7 × 6)/(5 × 4 × 3 × 2 × 1) 
= 252

Clearly,
n(S) = 6 × 6
= 36.
Let,
E be the event that the sum of the numbers on the two faces is divisible by 4 or 6.
Then, E = {(1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (5, 1), (5, 3), (6, 2), (6, 6)}
∴ n(E) = n(E)/n(S)
= 14/36
= 7/18.

Question: A football team has 15 players. In how many ways can a team of 11 players be chosen if the goalkeeper must always be included?

সমাধান:
15 জন খেলোয়াড় থেকে 1 জনকে (গোলরক্ষক) ঠিক রেখে বাকি 14 জন থেকে (11 - 1) = 10 জনের টিম গঠন করা যাবে।

∴ অবশিষ্ট 14 জন থেকে 10 জনকে নির্বাচন করার উপায় = ¹⁴C₁₀
= 14!/(14 - 10)! × 10!
= (14 × 13 × 12 × 11 × 10!)/(4! × 10!)
= (14 × 13 × 12 × 11)/(4 × 3 × 2 × 1)
= 1001

∴ 1001 উপায়ে দল গঠন করা যাবে।

Question: A box contains 12 red balls, 5 black balls, and 8 white balls. One ball is drawn at random. What is the probability that the ball drawn is neither red nor white?

Solution:
Number of red balls = 12
Number of black balls = 5
Number of white balls = 8

∴ Total balls = 12 + 5 + 8 = 25

Let event E = The ball drawn is neither red nor white, so it must be black.

∴ Number of favorable outcomes = 5
∴ P(E) = 5/25 = 1/5

Question: In how many ways can the letters of the word "EQUATION" be arranged such that the consonants occupy only the even positions?

Solution:
এখানে "EQUATION" শব্দটিতে মোট বর্ণ আছে 8টি।
ব্যঞ্জনবর্ণ (Consonant) আছে 3টি: Q, T, N
স্বরবর্ণ (Vowel) আছে 5টি: E, U, A, I, O
8টি বর্ণের মধ্যে জোড় স্থান (Even positions) আছে 4টি (2nd, 4th, 6th এবং 8th)।

4টি জোড় স্থানের মধ্যে 3টি ব্যঞ্জনবর্ণ সাজানোর উপায় = ⁴P₃ = 24
বাকি (8 - 3) = 5টি স্থানে 5টি স্বরবর্ণ সাজানোর উপায় = 5! = 120

∴ ব্যঞ্জনবর্ণগুলোকে কেবল জোড় স্থানে রেখে মোট বিন্যাস সংখ্যা = 24 × 120
= 2,880

অতএব, EQUATION শব্দটির ব্যঞ্জনবর্ণগুলোকে জোড় স্থানে রেখে মোট 2,880 উপায়ে সাজানো যাবে।

Here,
Total results 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
= 4.5/10
= 0.45
Answer: 0.45

Here S = {TTT, TTH,THT, HTT, THH, HTH, HHT, HHH}.
Let,
E = event of getting at least two heads
= {THH, HTH, HHT, HHH}.
∴ P(E) = n(E)/n(S)
= 4/8
= 1/2.

Let,
S be the sample space and
E be the event of select.
Then,
n(S) = number of ways of selecting 3 students out of 25
= ²⁵C₃
= (25 × 24 × 23)/(3 × 2 × 1)
= 2300.
And,
n(E) = (¹⁵C₂ × ¹⁰C₁)
= {(15 × 14)/(2 × 1)} × 10
= 1050.
∴ P(E) = n(E)/n(S)
= 1050/2300
= 21/46.

Question: In how many different ways can the letters of the word 'SOFTWARE' be arranged in such a way that the vowels always come together?

Solution:
The given word contains 8 different letters.
We keep the vowels (OAE) together and treat them as 1 letter.
Thus, we have to arrange the 6 letters SFTWR(OAE)
These can be arranged in 6! = 6 5 × 4 × 3 × 2 × 1 = 720 ways

And,
The vowels (OAE) can be arranged among themselves in 3! = 3 × 2 × 1 = 6 ways.

∴ Required number of ways = (720 × 6) = 4320

Question: Two dice are thrown together. What is the probability that the sum of the numbers on the two faces is divisible by 4 or 6 ?

Solution:
Two fair dice are thrown together.
So total possible outcomes = 6 × 6 = 36

And, 
Let E be the event that the sum of the numbers on the two faces is divisible by 4 or 6.
Then E = {(1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (5, 1), (5,  3), (6, 2), (6, 6)}
∴ n (E) = 14.

Hence, P(E) = n(E)/n(S) = 14/36
= 7/18

So the probability that the sum is divisible by 4 or 6 is 7/18.

Question: There are 3 doors to a lecture room. In how many ways can a lecturer enter the room from one door and leave from another door?

Solution: 
As the lecturer can't leave the hall by the door he/she enters.

So, number of ways can a student enter the hall through a door and leave the hall by a different door is = 3 × 2 = 6

Question: What is the value of (¹⁰P₁ × ⁵P₃).

Solution:
¹⁰P₁ = 10!/(10 - 1)!
= 10!/9!
= (10 × 9!)/9! 
= 10

⁵P₃ = 5!/(5 - 3)! 
= 5!/2!
= (5 × 4 × 3 × 2!)/2!
= 60

∴ ¹⁰P₁ × ⁵P₃ = 10 × 60 
= 600  

Let the 3 boys be B1, B2, B3 and 4 prizes be P1, P2, P3 and P4
Now B1 is eligible to receive any of the 4 available prizes (so 4 ways)
B2 will receive prize from rest 3 available prizes(so 3 ways)
B3 will receive his prize from the rest 2 prizes available(so 2 ways)
So total ways would be: 4 × 3 × 2 × 1 = 24 Ways
Hence, the 4 prizes can be distributed in 24 ways