ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

The given word is FORMULATE
The given word contains 9 letters, all different.
∴ Required number of ways =⁹P₉
                                              =9!
                                              =(9 × 8 × 7× 6 × 5 × 4 × 3 × 2 × 1)
                                              =362880

When three coins are tossed, total possible outcomes = 8
S = {HHH, HHT, HTT, THH, TTH, THT, HTH,TTT}
Favorable cases = {HHT,HTT,THH,TTH,THT,HTH} 

P(getting at least one head, one tail) = 6/8
                                                           = 3/4
∴ The probability is 3/4

The word 'RUMOUR' has two repetitive R and U
Therefore , number of arrangements = 6!​/(2! × 2!)
                                                            =180

When a dice is thrown once.
The total number of outcomes is 6 (1, 2, 3, 4, 5, and 6)
Odd numbers = 3 (1, 3, 5)

Probability = No of Favorable Outcomes/Total no of Outcomes
P(Odd numbers) = 3/6
                           = 1/2

∴ The required probability is 1/2

Given that 
Men = 8 
Women = 10

Required number of ways =⁸C₅ × ¹⁰C₆
                                          = 56 × 210
                                          = 11,760

Here,
n(S) = 52
Let E = event of getting a queen of club or a king of heart.
Then, n(E)= 2

P(E)=n(E)/​n(S)
     = 2/52
     = 1/26

The word 'OPTICAL' contains 7 different letters.
When the vowels OIA are always together, they can be supposed to form one letter.
Then, we have to arrange the letters PTCL (OIA).
Now, 5 letters can be arranged in 5! ways 
                                                   = 120 ways.
The vowels (OIA) can be arranged among themselves in 3! = 6 ways.

∴ Required number of ways = (120 x 6) = 720

The word 'EXTRA' contains 5 different letters.
The word EXTRA can be arranged in 5! ways = 120 ways

When the vowels EA are always together, they can be supposed to form one letter.
Then, we have to arrange the letters XTR(EA).
Now, 4 letters can be arranged in 4! ways 
                                                   = 24 ways.
The vowels (EA) can be arranged among themselves in 2! = 2 ways.
∴ The word EXTRA can be arranged in such a way that the vowels will be together = (24 x 2) =48


The letters of the words EXTRA be arranged so that the vowels are never together = (120 - 48) = 72 ways.

Total number of marbles = (2 + 4 + 3) = 9
Let E be the event of drawing 3 marbles such that none is blue.
Then, n (E) = number of ways of drawing 3 marbles out of 5 = ⁵C₃ = 10
 n(S)=⁹C₃
       =  84
∴P(E)=n(E)/n(S)
          = 10/84
          = 5/42

∴ Required probability
  = 1 - P(E)
  = 1 - (5/42)
  = (42 - 5)/42
  = 37/42

Men can be treated as one group.
So, five boys and 1 group of men can be arranged in 6! ways.
Two men can be arranged in 2! ways.
Total no. of cases = 6! × 2!
                            = 720 × 2
                            = 1440 ways

∴ Required no. of ways =1440

Number of ways of selecting (3 consonants out of 6) and (2 vowels out of 3)
= ⁶C₃ × ³C₂
= 20 × 3
= 60

Number of coins tossed = 3
∴ Sample space of three coins tossed = 2³ = 8

The different combination of men and women to form the group
3 men and 1 woman + 2 men and 2 women + 3 women and 1 man + 4 women
The selection of required man and women from 8 men and 6 women
⇒ ⁸C₃ × ⁶C₁ + ⁸C₂ × ⁶C₂ + ⁸C₁ × ⁶C₃ + ⁶C₄
⇒ 56 × 6 + 28 × 15 + 8 × 20 + 15
⇒ 336 + 420 + 160 + 15 
⇒ 931

Given word Punctual 

The word PUNCTUAL consists of 8 letters in total.
The word has five consonants - p, n, c, t, l
The word has three vowels -a, u and u -  where the letter ‘U’ comes twice.

The number of arrangements for the said word will be
= 8! / 2! 
= (8 × 7 × 5 × 4 × 3 × 2 × 1)/(1 × 2)
= 20160

Here S= {HH,HT,TH,TT}
Let E= event of getting at least on head ={HT,TH,TT}

∴P(E)=n(E)/n(S)​=3/​4

Number of handshakes in a conference = n(n - 1)​/2 where n being the number of people.

Given, there  are 9 people present in the conference.
So, number of handshakes in the conference = 9(9 - 1)​/2 = 36

Total outcome = 15 + 25 = 40 
Favorable outcome = 15
P (getting a prize) =15/40
                              = 3/8

Required number of ways
=(³C₁ × ⁵C₂)
= 3 × 10
= 30

Given that
Number of black balls = 4
Number of white balls = 6

Favorable event = ⁶C₂ = 15
Total possible events = ¹⁰C₂ = 45
∴ Probability = 15/45
                       = 1/3

Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}

Let E = event of getting at most two tails.
Then E = {TTH, THT, HTT, THH, HTH, HHT, HHH}

P(E) = n(E)/n(S)
     = 7/8

Ways in which at least 3 men are selected;
⇒ 3 men + 2 women
⇒ 4 men + 1 woman
⇒ 5 men + 0 woman

Number of ways = ⁷C₃ × ⁶C₂ + ⁷C₄ × ⁶C₁ + ⁷C₅ × ⁶C₀
                           = 35 × 15 + 35  × 6 + 21
                           = 756
                     
∴ The required no of ways = 756

Number of possible outcomes when two dice are thrown simultaneously: 6 × 6 = 36
(1,1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2),
(4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)

The possible outcomes =(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)
Hence, number of possible outcomes = 6

Required probability =6/36
                                 = 1/6