All questions of Permutation, Combination & Probability for Bank Exams Exam

For any multiple independent event, there are n^m

total possible outcomes, where n is the number of outcomes per event, and m is the number of such events.

So for a coin, discounting the unlikely event of landing on its side, there are two possible outcomes per event, heads or tails. And it is stated that there are 3 such events. So n^m=2³=8

.

Let us find the one pair of values for a, b.
a = 4, b = 19 satisfies this equation.
2*4 + 5*19 = 103.

Now, if we increase ‘a’ by 5 and decrease ‘b’ by 2 we should get the next set of numbers. We can keep repeating this to get all values.
Let us think about why we increase ‘a’ by 5 and decrease b by 2.
a = 4, b = 19 works.

Let us say, we increase ‘a’ by n, then the increase would be 2n.
This has to be offset by a corresponding decrease in b.
Let us say we decrease b by ‘m’.
This would result in a net drop of 5m.
In order for the total to be same, 2n should be equal to 5m.
The smallest value of m, n for this to work would be 2, 5.

a = 4, b = 19
a = 9, b = 17
a = 14, b = 15
..
And so on till
a = 49, b = 1
We are also told that ‘a’ should be greater than ‘b’, then we have all combinations from (19, 13) … (49, 1).
7 pairs totally.

Hence the answer is "7"

Choice C is the correct answer.

Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.
The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.
The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.
∴ Required number of numbers = (1 x 5 x 4) = 20.

The number of arrangements of the word ENGINEERING is 277200.
ENGINEERING word has 3 times of 3, three times of N, 2 times of G and 2 times of I. Then, the total letter is 11.

So, the number of arrangements of the word ENGINEERING = 11!/[3! * 3! * 2! * 2!] = 39916800/[6 * 6 * 2 * 2] = 277200

Hence, the number of arrangements of the word ENGINEERING is 277200.

In the word 'MATHEMATICS', we'll consider all the vowels AEAI together as one letter.

Thus, we have MTHMTCS (AEAI).

Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice

 Number of ways of arranging these letters =8! / ((2!)(2!))= 10080.

Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.

Number of ways of arranging these letters =4! / 2!= 12.

 Required number of words = (10080 x 12) = 120960

The bag contains 2 White, 3 Black and 4 Red balls.

So, total 9 balls are there in the bag; among them 3 are Black and 6 are non-Black balls.

Three balls can randomly be drawn in (9C3) = 84 ways.

1 Black and 2 non-Black balls can be drawn in (3C1)*(6C2) = 45 ways.

1 non-Black and 2 Black balls can be drawn in (6C1)*(3C2) = 18 ways.

3 Black balls can be drawn in (3C3) = 1 way.

So, three balls drawn in (45 + 18 + 1) = 64 ways will have at least one Black ball among the drawn ones.

The word 'LOGARITHMS' has 10 different letters.

Hence, the number of 3-letter words(with or without meaning) formed by using these letters 

= 10P3

= 10 * 9 * 8

= 720

There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants.

Let us mark these positions as under: 

                                                      (1) (2) (3) (4) (5) (6) 

Now, 3 vowels can be placed at any of the three places out 4, marked 1, 3, 5.  

Number of ways of arranging the vowels = 3P3

 = 3! = 6.

Also, the 3 consonants can be arranged at the remaining 3 positions. 

Number of ways of these arrangements = 3P3

 = 3! = 6. 

Total number of ways = (6 x 6) = 36.

The given word contains 5 different letters.
Keeping the vowels UE together, we suppose them as 1 letter.
Then, we have to arrange the letters JDG (UE).
Now, we have to arrange in 4! = 24 ways.
The vowels (UE) can be arranged among themselves in 2 ways.
∴ Required number of ways = (24 × 2) = 48

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 * 6) = 720.

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!=120 ways. The vowels (OIA) can be arranged among themselves in 3!=6 ways. Required number of ways =(120∗6)=720.

Vowels in the word "CORPORATION" are O,O,A,I,O

Lets make it as CRPRTN(OOAIO)

This has 7 lettes, where R is twice so value = 7!/2!

= 2520

Vowel O is 3 times, so vowels can be arranged = 5!/3!

= 20

Total number of words = 2520 * 20 = 50400

We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys). 

Required number of ways = (6C1*4C3)+(6C2*4C2)+(6C3*4C1)+6C4  

= (6C1*4C1)+(6C2*4C2)+(6C3*4C1)+6C2 = 209.

Explanation :

The word 'DELHI' has 5 letters and all these letters are different.

Total words (with or without meaning) formed by using all these
5 letters using each letter exactly once
= Number of arrangements of 5 letters taken all at a time
= ⁵P₅ = 5! = 5 x 4 x 3 x 2 x 1 = 120

Number should be a multiple of 3 and 4. So, the sum of the digits should be a multiple of 3. We can either have all seven digits as 3, or have three 2's and four 3's, or six 2's and a 3.
(The number of 2's should be a multiple of 3).

For the number to be a multiple of 4, the last 2 digits should be 32. Now, let us combine these two.
All seven 3's - No possibility.

Three 2's and four 3's - The first 5 digits should have two 2's and three 3's in some order.
No of possibilities = 5!3!2!5!3!2! = 10

Six 2's and one 3 - The first 5 digits should all be 2's. So, there is only one number 2222232.
So, there are a total of 10 + 1 = 11 solutions.

Hence the answer is "11"

Choice D is the correct answer.

Three-digit numbers: A B C. 3 can be printed in the 100’s place or 10’s place or units place.

=> 100’s place: 3 B C. B can take values 0 to 9, C can take values 0 to 9. So, 3 gets printed in the 100’s place 100 times
=> 10’s place: A 3 C. A can take values 1 to 9, C can take values 0 to 9. So, 3 gets printed in the 10’s place 90 times
=> Unit’s place: A B 3. A can take values 1 to 9, B can take values 0 to 9. So, 3 gets printed in the unit’s place 90 times

So, 3 gets printed 280 times in 3-digit numbers
Four-digit numbers: A B C D. 3 can be printed in the 1000’s place, 100’s place or 10’s place or units place.
=> 1000’s place: 3 B C D. B can take values 0 to 9, C can take values 0 to 9, D can take values 0 to 9. So, 3 gets printed in the 100’s place 1000 times.
=> 100’s place: A 3 C D. A can take values 1 to 9, C & D can take values 0 to 9. So, 3 gets printed in the 100’s place 900 times.
=> 10’s place: A B 3 D. A can take values 1 to 9, B & D can take values 0 to 9. So, 3 gets printed in the 10’s place 900 times.
=> Unit’s place: A B C 3. A can take values 1 to 9, B & C can take values 0 to 9. So, 3 gets printed in the unit’s place 900 times.

3 gets printed 3700 times in 4-digit numbers.
So, there are totally 3700 + 280 = 3980 numbers.

Hence the answer is "3980", Choice C is the correct answer.