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Test Level 1: Permutation & Combination - 2 - CAT MCQ


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20 Questions MCQ Test - Test Level 1: Permutation & Combination - 2

Test Level 1: Permutation & Combination - 2 for CAT 2024 is part of CAT preparation. The Test Level 1: Permutation & Combination - 2 questions and answers have been prepared according to the CAT exam syllabus.The Test Level 1: Permutation & Combination - 2 MCQs are made for CAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test Level 1: Permutation & Combination - 2 below.
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Test Level 1: Permutation & Combination - 2 - Question 1

The number of squares in a board having 10 rows and 9 columns is

Detailed Solution for Test Level 1: Permutation & Combination - 2 - Question 1

Number of squares in a board having m rows and n columns (m ≥ n) = m × n + (m - 1) × (n - 1) + (m - 2) × (n - 2) ... (m - n + 1) × 1
Here, m = 10 and n = 9
Number of squares = 10 × 9 + 9 × 8 + 8 × 7 + 7 × 6 + 6 × 5 + 5 × 4 + 4 × 3 + 3 × 2 + 2 × 1
= 90 + 72 + 56 + 42 + 30 + 20 + 12 + 6 + 2 = 330

Test Level 1: Permutation & Combination - 2 - Question 2

Find the last digit of the sum 1! + 2! + 3! + ………… + 100!.  

Detailed Solution for Test Level 1: Permutation & Combination - 2 - Question 2

1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
6! = 6 x 5! = shall always have 0 in the unit's place.
And the same is the case for 7!, 8! and the succeeding factorials.
Hence, unit's digit shall be the unit's digit in 1 + 2 + 6 + 24 + 0 = 3.

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Test Level 1: Permutation & Combination - 2 - Question 3

Two packs of 52 cards are shuffled together. The number of ways in which a man can be dealt 26 cards, so that he does not get two cards of the same suit and same denominations is

Detailed Solution for Test Level 1: Permutation & Combination - 2 - Question 3

26 cards can be chosen out of 52 cards in 52C26 ways. There are two ways in which each card can be dealt, either from the first pack or from the second.
Total number of ways = 52C26 . 226

Test Level 1: Permutation & Combination - 2 - Question 4

In how many ways will a particular person be found among those six if six persons are to be selected out of ten?

Detailed Solution for Test Level 1: Permutation & Combination - 2 - Question 4

We require that the particular person is found in every group of 6 people out of a total of 10. As this person is common to all groups of 6, we will have to consider the number of ways of choosing the remaining 5 persons out of the other 9. This can be done in 9C5 ways = 126 ways.

Test Level 1: Permutation & Combination - 2 - Question 5

A 7-character code is such that only even digits occur at even places and only odd digits occur at odd places, e.g. 1436789. How many such codes can be made from digits 1-9, if repetition of digits is allowed?

Detailed Solution for Test Level 1: Permutation & Combination - 2 - Question 5

Odd digits: 1, 3, 5, 7, and 9
Even digits: 2, 4, 6, and 8
Number of ways to fill an odd place = 5
Number of ways to fill an even place = 4

Therefore,
Total number of cases = (5)4 × (4)3

Test Level 1: Permutation & Combination - 2 - Question 6

Five boxes are numbered from 1 to 5. A ball is to be placed in each box and balls of three different colours Red, Blue and Green are available. After this, a code is generated as *****, where the first * represents the first letter of colour of the ball in that box, and so on so forth (e.g. RBBGR and RBBRG are two different codes). How many different codes will result?

Detailed Solution for Test Level 1: Permutation & Combination - 2 - Question 6

For every box, we have three different possibilities of colours.
Therefore, total number of possibilities = 3 × 3 × 3 × 3 × 3

Test Level 1: Permutation & Combination - 2 - Question 7

If 10P5 × 4P3 = 6P5 × X × 2, then X is

Detailed Solution for Test Level 1: Permutation & Combination - 2 - Question 7

10P5 × 4P3 = 6P5 × X × 2
10 × 9 × 8 × 7 × 6 × 4 × 3 × 2 = 6 × 5 × 4 × 3 × 2 × X × 2
9 × 8 × 7 = X then X = 9P3

Test Level 1: Permutation & Combination - 2 - Question 8

What is the difference between total number of arrangements when 10 members are arranged in a row and in a circle?

Detailed Solution for Test Level 1: Permutation & Combination - 2 - Question 8

Total number of arrangements when n members are arranged in a row = n!
Total number of arrangements when n members are arranged in a circle = (n - 1)!
Difference = n! - (n - 1)!
Difference = 10! - 9!

Test Level 1: Permutation & Combination - 2 - Question 9

How many numbers lying between 600 and 700 can be formed with the digits 0-9, where 7 occurs at exactly one of the positions (repetition is allowed)?  

Detailed Solution for Test Level 1: Permutation & Combination - 2 - Question 9

Let 7 occur in the units place. So, the units digit can be placed in 1 way. The tens digit can be placed by the rest of 9 numbers. The hundreds digit can be placed by 6 only, i.e. one way.
As the numbers are between 600 and 700, number of ways = 1 × 9 × 1 = 9
Now, let 7 occur in the tens place. The units place can be filled by any of the numbers from 0-9 except 7.
The hundreds place can be filled in one way (by only 6). So, total 9 ways. 7 can't occur in the hundreds place.
∴ Total number of ways in which numbers can be formed = 9 + 9 = 18

Test Level 1: Permutation & Combination - 2 - Question 10

In how many ways can 6 awards be distributed among 9 actors, if each actor can get any number of awards?

Detailed Solution for Test Level 1: Permutation & Combination - 2 - Question 10

For 1st award: We have 9 options to distribute.
And for 2nd award: Again we have 9 options to distribute (Any actor can get any number of award)
Similarly for each others we have 9 options.
Total number of ways when repetition is allowed = 9 × 9 × 9 × 9 × 9 × 9 = 96
So, total number of ways of distributing the awards = 96

Test Level 1: Permutation & Combination - 2 - Question 11

There are 6 black balls and 4 white balls in a box. In how many ways can 4 balls be selected, in which there is at least one white ball?  

Detailed Solution for Test Level 1: Permutation & Combination - 2 - Question 11

Selecting 4 balls out of 10 balls having at least 1 white ball = Selecting 4 balls out of 10 balls - Selecting 4 balls out of 10 balls having no white ball
= 10C4 - 6C4   

= 210 - 15 = 195

Test Level 1: Permutation & Combination - 2 - Question 12

An apparel shop has 15 different shades of T-shirts and 35 different shades of shorts. In how many different ways can one buy two items containing only one shade of T-shirt and one shade of shorts?  

Detailed Solution for Test Level 1: Permutation & Combination - 2 - Question 12

For every single shade of a T-shirt, one can make 35 distinct pairs with differently shaded shorts. Since there are 15 shades of T-shirts, number of ways = 15 × 35 = 525

Test Level 1: Permutation & Combination - 2 - Question 13

How many numbers of 3 digits can be formed with the digits 1, 2, 3, 4, 5 (repetition of digits not allowed)?

Detailed Solution for Test Level 1: Permutation & Combination - 2 - Question 13

The correct option is C 60
Three digits can be formed in
5P3 ways that is
5!/(5-3)! = 60

Test Level 1: Permutation & Combination - 2 - Question 14

In how many ways can a person send invitation cards to 6 of his friends if he has four servants to distribute the cards?

Detailed Solution for Test Level 1: Permutation & Combination - 2 - Question 14

Each invitation card can be sent in 4 ways. Thus, 4 × 4 × 4 × 4 × 4 × 4 = 46.

Test Level 1: Permutation & Combination - 2 - Question 15

In how many ways can 7 Indians, 5 Pakistanis and 6 Dutch be seated in a row so that all persons of the same nationality sit together?

Detailed Solution for Test Level 1: Permutation & Combination - 2 - Question 15

We need to assume that the 7 Indians are 1 person, so also for the 6 Dutch and the 5 Pakistanis. These 3 groups of people can be arranged amongst themselves in 3! ways. Also, within themselves the 7 Indians the 6 Dutch and the 5 Pakistanis can be arranged in 7!, 6! And 5! ways respectively. Thus, the answer is 3! × 7! × 6! × 5!.

Test Level 1: Permutation & Combination - 2 - Question 16

There are 4 qualifying examinations to enter into Oxford University: RAT, BAT, SAT, and PAT. An Engineer cannot go to Oxford University through BAT or SAT. A CA on the other hand can go to the Oxford University through the RAT, BAT & PAT but not through SAT. Further there are 3 ways to become a CA(viz., Foundation, Inter & Final). Find the ratio of number of ways in which an Engineer can make it to Oxford University to the number of ways a CA can make it to Oxford University.

Detailed Solution for Test Level 1: Permutation & Combination - 2 - Question 16

An engineer can make it through in 2 ways, while a CA can make it through in 3 ways. Required ratio is 2:3. Option (b) is correct.

Test Level 1: Permutation & Combination - 2 - Question 17

If nC3 = nC8, find n.

Detailed Solution for Test Level 1: Permutation & Combination - 2 - Question 17

Use the property nCr = nCn-r to see that the two values would be equal at n = 11 since 11C3 = 11C8.

Test Level 1: Permutation & Combination - 2 - Question 18

In how many ways can the letters of the word PATNA be rearranged?

Detailed Solution for Test Level 1: Permutation & Combination - 2 - Question 18

Rearrangements do not count the original arrangements. Thus, 5!/2! – 1 = 59 ways of rearranging the letters of PATNA.

Test Level 1: Permutation & Combination - 2 - Question 19

If 10Pr = 720, find r.

Detailed Solution for Test Level 1: Permutation & Combination - 2 - Question 19

10P3 would satisfy the value given as 10P3 = 10 × 9 × 8 = 720.

Test Level 1: Permutation & Combination - 2 - Question 20

How many numbers of four digits can be formed with the digits 0, 1, 2, 3 (repetition of digits being allowed?

Detailed Solution for Test Level 1: Permutation & Combination - 2 - Question 20

3 × 4 × 4 × 4 = 192

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