ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

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

Let S be the sample space
Then,
n(S) = ⁵²C₂
= 52 × 51/2 × 1
= 1326
Let E = event of getting 2 kings out of 4.
n(E) = 4C2
=> 4 × 3/2 × 1
P(E) = n(E)/n(S)
= 6/1326
= 1/221

P(first letter is not vowel) = 2/4
P(second letter is not vowel) = 1/3
So, probability that none of letters would be vowels is = 2/4 × 1/3
= 1/6

Here S = {1, 2, 3, 4, 5, 6}
Let E be the event of getting the multiple of 3
Then,
E = {3,6}
P(E) = n(E)/n(S)
= 2/6
= 1/3

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 x 6)/252
= 5/21

n(S) = ²⁰C₂ = 190
n(E) = ¹⁵C₂ = 105
Therefore,
P(E) = 105/190
= 21/38

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4) = (7C3 × 4C2)
= 210.
Number of groups, each having 3 consonants and 2 vowels = 210.
Each group contains 5 letters.
Number of ways of arranging 5 letters among themselves = 5! = 120
Required number of ways = (210 x 120)
= 25200.

The word 'LEADING' has 7 different letters.
When the vowels EAI are always together, they can be supposed to form one letter.
Then, we have to arrange the letters LNDG (EAI).
Now, 5 (4 + 1) letters can be arranged in 5! = 120 ways.
The vowels (EAI) can be arranged among themselves in 3! = 6 ways.
Required number of ways = (120 x 6)
= 720.

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

A team of 6 members has to be selected from the 10 players.
This can be done in 10C6 or 210 ways.
Now, the captain can be selected from these 6 players in 6 ways.
Therefore,
total ways the selection can be made is = 210 × 6
= 1260

We are to choose 11 players including 1 wicket keeper and 4 bowlers or 1 wicket keeper and 5 bowlers.
Number of ways of selecting 1 wicket keeper, 4 bowlers and 6 other players in ²C₁ × ⁵C₄ × ⁹C₆ = 840
Number of ways of selecting 1 wicket keeper, 5 bowlers and 5 other players in ²C₁ × ⁵C₅ × ⁹C₅ = 252
Total number of ways of selecting the team = 840 + 252 = 1092

There are 7 digits 1, 2, 0, 2, 4, 2, 4 in which 2 occurs 3 times, 4 occurs 2 times.
Number of 7 digit numbers = 7!3! × 2! = 420
But out of these 420 numbers,
there are some numbers which begin with '0' and they are not 7-digit numbers. The number of such numbers beginning with '0'.
= 6!3! × 2! = 60
Hence the required number of 7 digits numbers = 420 - 60 = 360

There are 15 dots in total,and to make a triangle we need to select any three of those dots.
So, ¹⁵C₃ = 455

Parallelograms are formed when any two pairs of parallel lines (where each pair is not parallel to the other pair) intersect.
Hence, the given problem can be considered as selecting pairs of lines from the given 2 sets of parallel lines.
Therefore, the total number of parallelograms formed = ⁷C₂ x ⁶C₂ = 315

Number of ways of choosing 2 black pens from 5 black pens in ⁵C₂ ways.
Number of ways of choosing 2 white pens from 3 white pens in ³C₂ ways.
Number of ways of choosing 2 red pens from 4 red pens in ⁴C₂ ways.
By the Counting Principle, 2 black pens, 2 white pens, and 2 red pens can be chosen in 10 x 3 x 6 = 180 ways

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

First letter can be posted in 4 letter boxes in 4 ways.
Similarly the second letter can be posted in 4 letter boxes in 4 ways and so on.
Hence all the 5 letters can be posted in = 4 x 4 x 4 x 4 x 4 = 1024

As the toys are distinct and not identical,
For each of the 8 toys, we have three choices as to which child will receive the toy.
Therefore, there are 5⁸ ways to distribute the toys.
Hence, it is 5⁸ and not 8⁵.

There are 7 letters in the word DRAUGHT, the two vowels are A and U.
Since, the vowels are not to be separated; AU can be considered as one entity.
Therefore, the number of letters will be 6 instead of 7.
The permutations will be P(6,6) = 6! ways.
But the two vowels A and U can be arranged in two ways, i.e. AU and UA.
The required number of arrangements = 2!.6! = 1440 ways.

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