Permutation & Combination MCQs for UPPSC (UP) Exam

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.

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 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

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.

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.