ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

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  

Question: If ⁿC₇ = ⁿC₅ , then find the value of n.

Solution:
We know, 
If ⁿC_r = ⁿC_s , then, n = r + s

Now, ⁿC₇ = ⁿC₅ 
∴ n = 7 + 5
= 12

Question: What is the probability of getting a sum of 9 when a die is thrown twice?

Solution: 
In two throws of a dice, n(S) = (6 x 6) = 36
Let E = the event of getting a sum
= {(3, 6), (4, 5), (5, 4), (6, 3)}
∴ n(E) = 4

Hence, P(E) = n(E)/n(S)
= 4/36
= 1/9

Question: In how many ways can 8 football players be divided into two teams with an equal number of players?

Solution:
8 জন খেলোয়ার কে সমান দুটি ভাগে ভাগ করলে 8/2 = 4 জন করে দল গঠন করা যাবে।
 
তাহলে, 8 জন থেকে 4 জন করে নিয়ে পাই,  ⁸C₄ = 8!/[4!(8 - 4)!] 
= (8 × 7 × 6 × 5 × 4!)/(4! × 4!)
= (8 × 7 × 6 × 5)/(4 × 3 × 2)
= 70 

∴ দল গঠন করার উপায় = 70/2 = 35

Question: In a lottery, there are 8 prizes and 12 blanks. A lottery is drawn at random. What is the probability of getting a prize?

Solution: 
Here, 
Number of prizes = 8
Number of blanks = 12
∴ Total outcome = 8 + 12
= 20

∴ Probability of getting a prize = Number of prizes/Total outcome 
= 8/20
= 2/5 

Question: In how many ways can the letters of the word 'TRIANGLE' be arranged, if the vowels always stay together?

Solution: 
Triangle শব্দটিতে মোট বর্ণ  8 টি (সবই ভিন্ন) এবং স্বরবর্ণ 3 টি।
যেহেতু, স্বরবর্ণ 3 টি একত্রে থাকবে, সেহেতু স্বরবর্ণ তিনটিকে একটি বর্ণ ধরে মোট বর্ণ সংখ্যা 6 টি। 
এই 6 টি বর্ণকে নিজেদের মধ্যে সাজানো যায় = 6! উপায়ে
আবার, স্বরবর্ণ  3 টিকে নিজেদের মধ্যে সাজানো যায় = 3! উপায়ে

∴ সর্বমোট বিন্যাস = 6! × 3!
= 4320

Question: The ratio of the number of red balls to yellow balls to green balls in an urn is 3 : 4 : 5. What is the probability that a ball chosen at random from the urn is a red ball? 

Solution:
লাল, হলুদ, ও সবুজ বলের সংখ্যার আনুপাতিক মান যথাক্রমে 3, 4, 5
মোট বলের সংখ্যার আনুপাতিক মান = 3 + 4 + 5
 = 12

∴ বলটি লাল হওয়ার সম্ভাব্যতা = 3/12
= 1/4

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 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: The probability that a card drawn from a pack of 52 cards will be a diamond or a king is: 

Solution: 
Here, n(S) = 52
There are 13 cards of diamonds (including one king), and there are three more kings.

Let E = the event of getting a diamond or a king
Then, n(E) = (13 + 3) = 16

∴ P(E) = n(E)/n(S)
= 16/52
= 4/13

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 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

Question: A bag contains 7 red and 9 green marbles. One marble is drawn at random. What is the probability that the marble drawn is not red?

Solution:
Number of red marbles = 7
Number of green marbles = 9
Total number of marbles = 7 + 9 = 16

P (red marble) = 7/16

∴ P (not red marble) = 1 - (7/16)
= 9/16

Question: If the chairperson's seat is fixed, in how many ways can 6 people be seated at a circular table?

Solution:
Chairperson’s seat is fixed
∴ Remaining 5 people can be arranged in 5! ways
= 5 × 4 × 3 × 2 
= 120 ways

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

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: 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: 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!

Question: In a survey, it was found that 70% of people read Ittefaq, 60% read Sangbad, and 40% read both newspapers. If a person is chosen at random, find the probability that they read either Ittefaq or Sangbad.

Solution:
ধরি, ইত্তেফাক পড়ার ঘটনা A 
এবং সংবাদ পড়ার ঘটনা B
P(A) = 70/100 = 7/10 
P(B) = 60/100 = 6/10
P(A ∩ B)= 40/100 = 4/10

নিরপেক্ষভাবে বাছাই করলে একজন লোকের ইত্তেফাক বা সংবাদ পড়ার সম্ভাব্যতা P( A ∪ B)

∴ P(A ∪ B) = P(A) + P(B) - P(A ∩ B) 
= (7/10) + (6/10) - (4/10)
= (7 + 6 - 4)/10
= 9/10

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 

Question: What is the probability of having 53 Sundays in a leap year?

Solution:
Leap year = 366 Days
= {(52 × 7) + 2} Days
= 52 Weeks + 2 Days

∴ Probability of having 53 Sundays = 2/7

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

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

Question: Find the value of (x + 1)!/(x - 2)! 

Solution:
Here,
(x + 1)!/(x - 2)!
= [x(x + 1)(x - 1)(x - 2)!]/(x - 2)! 
= x(x + 1)(x - 1)
= x(x² - 1) 

Question: In a class, there are 10 boys and 8 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected is:

Solution: 
Total students = 10 + 8 = 18
Let S be the sample space, and E be the event of selecting 1 girl and 2 boys.

Then, n(S) = Number of ways of selecting 3 students out of 18 = ¹⁸C₃
= (18 × 17 × 16)/(3 × 2 × 1)
= 816

And, n(E) = Number of events of selecting 1 girl and 2 boys = ⁸C₁ × ¹⁰C₂ 
= 8 × [(10 × 9)/2]
= 8 × 45
= 360

∴ Probability = n(E)/n(S)
= 360/816
= 15/34