CAT Practice: Permutation & Combination - 1 for UPSC - Free MCQ Test

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.

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.

Let us assume that the student answers all the questions.

If he has answered all of the questions correctly, he gets a total of 34 x 3 = 102 marks.

Now, for every wrong answer, there is a subtraction of 4 marks, so for n incorrect answers, marks achieved = 102 - 4n

Now, we need to equate this to 67, which is the actual marks achieved. But this is not possible, because 67 is odd, and 102-4n is even

Similarly, we can assume he had attempted 33 questions, with n wrong answers,

33 x 3 - 4n = 67

99 - 4n = 67

4n = 32, n = 8. This is possible.

For 32 attempted questions also,

32 x 3 - 4n = 67 is not possible because left side is even and right side is odd.

For 31 attempted questions,

31 x 3 - 4n = 67

93 - 4n = 67

4n = 26, not possible

Similarly, 30 attempted questions is not possible.

For 29 attempted questions,

29 x 3 - 4n = 67

87 - 4n = 67

n = 5. This is possible.

Again for 28, 27, 26 attempted questions, it is not possible.

For 25 questions attempted,

25 x 3 - 4n = 67

75 - 4n = 67

n = 2

It is not possible for any other combination.

Hence, it is possible in 3 different ways.

The domain of ⁿC_r is such that r > 0 and n > r.
So, x+3 ≥ 0  and 15-2x≥x+3 are the only two conditions which needs to be satisfied.
x > -3 and 12> 3x x > -3 x < 4  are the two conditions.
.'. The possible integer values of x are -3, -2, -1, 0, 1, 2, 3 and 4.

35 _ Can be chosen in 6 ways.

35 _ _ We can choose 2 out of the remaining 6 in ⁶C₂ = 15 ways. We remove 1 case where 7 and 8 are together to get 14 ways.

35 _ _ _We can choose 3 out of the remaining 6 in ⁶C₃ = 20 ways. We remove 4 cases where 7 and 8 are together to get 16 ways.

35 _ _ _ _We can choose 4 out of the remaining 6 in ⁶C₄ = 15 ways. We remove 6 case where 7 and 8 are together to get 9 ways.

35 _ _ _ _ _ We choose 1 out of 7 and 8 and all the remaining others in 2 ways.

Thus, total number of cases = 6+14+16+9+2 = 47.

Alternatively,

The arrangement requires a selection of 3 or more numbers while including 3 and 5 and 7, 8 are never included together. We have cases including a selection of only 7, only 8 and neither 7 nor 8.

Considering the cases, only 7 is selected.

We can select a maximum of 7 digit numbers. We must select 3, 5, and 7.

Hence we must have ( 3, 5, 7) for the remaining 4 numbers we have

Each of the numbers can either be selected or not selected and we have 4 numbers :

Hence we have _ _ _ _ and 2 possibilities for each and hence a total of 2*2*2*2 = 16 possibilities.

SImilarly, including only 8, we have 16 more possibilities.

Cases including neither 7 nor 8.

We must have 3 and 5 in the group but there must be no 7 and 8 in the group.

Hence we have 3 5 _ _ _ _.

For the 4 blanks, we can have 2 possibilities for either placing a number or not among 1, 2, 4, 6. = 16 possibilities But we must remove the case where neither of the 4 numbers are placed because the number becomes a two-digit number.

Hence 16 - 1 = 15 cases.

Total = 16+15+16 = 47 possibilities

Case 1 'zero vertices on the given line'

Since there are exactly three points not on the line, there can only be one triangle formed with these three points.

Case 2 'one vertex on the given line'

There are four choices for the point on the line and for each of these four points there are three ways of selecting the pair of vertices not on the line. Thus, there are 3 × 4 or 12 possible triangles

Case 3 '2 vertices on the given line'

There are six ways of choosing the pair of points on the line and for each of these six pairs there are three ways of selecting the vertex not on the line giving a total of 6 × 3 or 18 possibilities.

In total there are 1+12+18 or 31 triangles

So total possible triangles = 11C3 = 165.

However, AOB, COD, EOF, GOH, JOK lie on a straight line. Hence, these 5 triangles are not possible. Thus, the required number of triangles = 165 - 5 = 160

Single digit numbers with non-repeating digits = 9
(The unit’s digit is non-zero)
Two digit numbers with non-repeating digits = 9 × 9
(The tenth’s digit is non-zero and the unit digit can be any digit except the tenth’s digit.)
Three digit numbers with non-repeating digits = 9 × 9 × 8
(The hundred’s digit is non-zero and the tenth’s digit can be any digit except the hundred’s digit and the unit digit can be any digit except the tenth’s digit.)
So, totally there are (9 + 9 × 9 + 9 × 9 × 8) = 738 natural numbers up to 1000 with non-repeating digits.