ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

There are total 7 digits given - 2, 3, 4, 5, 6, 8 and 0
Between 500 and 1000 means from 501 to 999.

All of them are 3 digit numbers.
First digit - 5, 6 or 8 i.e. 3 possibilities (1 digit gets used here)
Second digit - Any one from 7-1 = 6 remaining digits i.e. 6 possibilities
Third digit - Any one from 6-1 = 5 remaining digits i.e. 5 possibilities

∴ Total numbers possible = 3 x 6 x 5 = 90.

'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.

There are 12 celebrities. A handshake needs 2 people.
This simply means in how many ways 2 people can be selected out of 12.

So the answer is ¹²C₂

ⁿC_r = n!/r!(n - r)!

∴ ¹²C₂ = 12!/2!(12 - 2)! = (12 × 11)/2
= 66 = number of handshakes. [If there are n people and they shake hands only once with each other, then, Number of handshakes = ⁿc₂ = n(n -1)/2]

We need to SELECT people.
[SELECT = Combination = ⁿC_r = n!/r!(n-r)!

There are 20 stations. A ticket is needed between 2 stops.
That means, we simply need to select 2 stops from possible 20 stops.

That can be done by ²⁰C₂ = 20!/2!(20 - 2)! = 20!/2!18! = 190 ways.

This is when we start from one side.
When we travel from the other side we will need a separate ticket.
That means while going from A to B and B to A, we will need separate tickets.
So again on another side, we need 190 tickets.

Total tickets = 190 + 190 = 380 tickets.

We need to SELECT people.
[SELECT = Combination = ⁿC_r = n!/r!(n-r)!

Here, we first have to select 10 ladies from 19.
Select = Combination
∴ Select 10 ladies = ¹⁹C₁₀

Arrange 10 ladies in circle = 10 - 1 = 9! ways
19 - 10 = 9 ladies remain.
Arrange the remaining 9 ladies in another circle = 9 - 1 = 8! ways

∴ Total ways to arrange 19 ladies in 2 required circles = ¹⁹C₁₀ × 9! × 8!

‘LOGARITHM’ contains 10 different letters.

Required number of word
= Number of arrangements of 10 letters, taking 4 at a time
= ¹⁰P₄ = (10 × 9 × 8 × 7) = 5040.

As per the question, three girls can’t occupy consecutive seats but two can.

Therefore, if we find the number of ways in which all three girls occupy consecutive seats and subtract this number from the total number of ways in which the five people can be arranged among themselves, we will get the required answer.

5 students can be arranged among themselves in ⁵p₅ ways = 120 ways.

Assume that the 3 girls are one entity. The total number of ways in which they can be arranged among themselves = 3! = 6
Also, the set of three girls and the other students can be arranged among themselves in 3! = 6 ways.
Thus, the total number of ways in which three girls are together = 6 × 6 = 36

Thus, a number of ways in which all 3 girls will not occupy consecutive seats = 120 – 36 = 84.

We know,
Probability = what we want/Total
Or = add; AND = multiply

We want black OR red pebble
There are 6 black and 8 red pebbles
Total pebbles = 2 + 6 + 4 + 8 = 20

So, Probability = 6/20 + 8/20 = 14/20 = 7/10.

We know,
Probability = what we want/Total
Or = add; AND = multiply

We want a grey card OR ENGLISH card
There are 30 - 17 = 13 grey cards
There are 4 + 5 = 9 ENGLISH cards
Total cards = 30
Also, 5 grey cards are ENGLISH cards.

So Probability = 13/30 + 9/30 - 5/30 = (13 + 9 - 5)/30
= 17/30 [This subtraction is needed a grey card gets counted twice - once in 13 grey cards and once again in 9 ENGLISH cards.]

We know,
Probability = what we want/Total
Or = add; AND = multiply

We want 2 orange towels
That means to choose one AND then choose other from the remaining towels
There are 6 orange towels
Total 15 towels

So probability for two orange towels = 6/15 × 5/14
= 1/7 [Here we reduce the denominator i.e. the total number of towels because once we remove a towel from the box we do not put it back in the box.
So while removing the 2^nd towel, there are only 15 - 1 = 14 towels in the box.]

We want 2 red and 1 blue shirt
There are 4 red shirts and 3 blue shirts
Total = 15 shirts
You can choose a blue shirt 1^st, then a red shirt and then a red shirt

Probability = 3/15 × 4/14 × 3/13 = 6/455

OR You can choose a red shirt 1^st, then redshirt and then a blue shirt
OR You can choose a red shirt 1^st, then a blue shirt and then red-shirt
For all 3 the probability remains the same = 6/455

We need to add these 3 probabilities to get the total probability
Total probability = 6/455 + 6/455 + 6/455 = 18/455.

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.]

A dice has 6 faces.
So there are 6 possible outcomes
Dice are rolled once AND then again.
So total possibilities = 6 x 6 = 36

The sum should be 8 of the 2 throws.
So which combination of numbers from 1 to 6 will yield us a sum of 8?
They are - (2,6); (6,2); (3,5); (5,3); (4,4)

So there are a total of 5 possibilities where the addition is 8.
But only 1 possibility where the first throw of dice is 4.

So, the Probability for the first throw to be 4 and sum to be 8 = 1/36.

When 4 fair coins are tossed simultaneously, the total number of outcomes is 2⁴ = 16
At least 3 heads imply that one can get either 3 heads or 4 heads.
One can get 3 heads in ⁴C₃ = 4 ways and can get 4 heads in ⁴C₄

∴ Total number of favorable outcomes = 4 + 1 = 5
∴ The required probability = 1/4.