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# Test: Permutation And Combination- 5

## 15 Questions MCQ Test General Aptitude for GATE 2020 | Test: Permutation And Combination- 5

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This mock test of Test: Permutation And Combination- 5 for UPSC helps you for every UPSC entrance exam. This contains 15 Multiple Choice Questions for UPSC Test: Permutation And Combination- 5 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Permutation And Combination- 5 quiz give you a good mix of easy questions and tough questions. UPSC students definitely take this Test: Permutation And Combination- 5 exercise for a better result in the exam. You can find other Test: Permutation And Combination- 5 extra questions, long questions & short questions for UPSC on EduRev as well by searching above.
QUESTION: 1

### In how many ways can the letters of the word DELHI be arranged?

Solution:

A and B can occupy the first and the ninth places, the second and the tenth places, the third and the eleventh place and so on... This can be done in 18 ways.
A and B can be arranged in 2 ways.
All the other 24 alphabets can be arranged in 24! ways.
Hence the required answer = 2 x 18 x 24!

QUESTION: 2

### How many numbers between 200 and 1200 can be formed with the digits 0, 1, 2, 3 (repetition of digits not allowed?

Solution:

14 numbers

So total of 14 numbers (without repetition) between 200 & 1200 with 0, 1, 2 & 3

QUESTION: 3
Solution:

28C2r/24C2r-4 = 225/11
⇒ 28!/2r!(28-2r)! * [(2r-4)!(28-2r)!]/24! = 225/11
⇒ (28*27*26*25)/[2r*(2r-1)*(2r-2)*(2r-3)] = 225/11
⇒ 2r*(2r-1)*(2r-2)*(2r-3) = (28*27*26*25*11)/225
⇒ 2r*(2r-1)*(2r-2)*(2r-3) = 28*3*26*11
⇒ 2r*(2r-1)*(2r-2)*(2r-3) = 4*7*3*13*2*11
⇒ 2r*(2r-1)*(2r-2)*(2r-3) = (2*7)*13*(3*4)*11
⇒ 2r*(2r-1)*(2r-2)*(2r-3) = 14*13*12*11
⇒ 2r = 14
⇒ r = 7

QUESTION: 4

In how many ways can 10 identical presents be distributed among 6 children so that each child gets at least one present?

Solution:

We have to count natural numbers which have a maximum of 4 digits. The required answer will be given by: Number of single digit numbers + Number of two digit numbers + Number of three digit numbers + Number of four digit numbers.

QUESTION: 5

A captain and a vice-captain are to be chosen out of a team having eleven players. How many ways are there to achieve this?

Solution:

Explaination: out of 11 player 1 captain can be choose 11 ways, Now remaining 10 player,wise captain can be choose in 10 ways Therefore total number of ways =11*10=110 ways

QUESTION: 6

In how many ways can Ram choose a vowel and a consonant from the letters of the word ALLAHABAD?

Solution:

only 1 vowel available for selection (A).
A is available 4 times .
there are 4 consonants available – L, H, B, D
Then the number of ways of selecting a vowel and a consonant would be 1 × 4C1 = 4.

QUESTION: 7

How many motor vehicle registration number of 4 digits can be formed with the digits 0, 1,2, 3, 4, 5? (No digit being repeated.)

Solution:

The correct option is Option D.

There are six different digits (0,1,2,3,4,5) and these will be arranged in 6p4 ways = 360

Hence, 360 different registration numbers can be formed with the help of these numbers.

QUESTION: 8

There are ten subjects in the school day at St.Vincent’s High School but the sixth standard students have only 5 periods in a day. In how many ways can we form a time table for the day for the sixth standard students if no subject is repeated?

Solution:

There are 10 subjects and 5 periods
I st period can be filled with any 10 subjects
2nd period can be filled with remaining 9 subjects (1 subject is already filled)
3 rd period can be filled with remaining 8 subjects (2 subjects are already filled)
4 th period can be filled with remaining 7 subjects (3 subjects are already filled)
5 rd period can be filled with remaining 6 subjects (4 subjects are already filled)
so total no fo ways we can arrange = 10*9*8*7*6
= 30240 ways

QUESTION: 9

How many batting orders are possible for the Indian cricket team if there is a squad of 15 to choose from such that Sachin Tendulkar is always chosen?

Solution:

The selection of the II player team can be done in 14C10 ways. This results in the team of 11 players be-ing completely chosen. The arrangements of these 11 players can be done in 11!.
Total batting orders = 14C10 x 11! = 1001 x 11!
(Note: Arrangement is required here because we are talking about forming batting orders).

QUESTION: 10

How many distinct words can be formed out of the word PROWLING which start with R & end with W?

Solution:
QUESTION: 11

How many even numbers of four digits can be formed with the digits 1, 2, 3, 4, 5, 6 (repetitions of digits are allowed)?

Solution:

for an even number, the ones place number should be even i.e it could be either one of 2,4,6

so number of ways to select ones place number is 3C1 = 3

since repetetion is allowed, and all numbers are whole numbers.

so number of ways to select any of the three remaining numbers = 6C1 =6

Number of 4 digit number possible = 6 x 6 x 6 x 3

= 648

QUESTION: 12

On a shelf, 2 books of Geology, 2 books of Sociology and 5 of Economics are to be arranged in such a way that the books of any subject are to be together. Find in how many ways can this be done?

Solution:

There are books of 3 subjects (Geology, Sociology and Economics), hence they can be arranged in 3! (3 * 2 * 1) = 6 ways.
Further, in each category (subject), books are to be arranged in different order, we get,
Required number of ways:

3! * [2! * 2! * 5!] = 2880

QUESTION: 13

In how many ways can the letters of the word ‘EQUATION’ be arranged so that all the vowels come together?

Solution:

The correct option is Option B.

The word equation has 8 letters of which 5 are vowels and 3 are consonants. Bunch up the 5 vowels to assume them as one letter. Thus the total number of letters is 4 which can be arranged in  4!=24 . And for each of these arrangements the 5 vowels can be arranged among themselves in  5!=120  ways. Thus, the total arrangements of the letters with the vowels always appearing together is  24∗120=2880 .

QUESTION: 14

How many quadrilateral can be formed from 25 points out of which 7 are collinear

Solution:

otal number of quadrilateral combination possible if none of the points are collinear = 25C4 = 12650

If we form a geometry by joining any three points out of seven collinear points and one point from 18 non collinear points, it will give us a triangle instead of quadrilateral. So we have to eliminate number of combinations which can be formed in this way, which is 7C3 x 18C1 = 35 x 18 = 630

We also can't form quadrilateral if we choose all four vertices of quadrilateral to be any 4 points from 7 collinear points. It will come out to be a straight line. So we have to eliminate such combinations also. Which is 7C4 = 35

So net number of possible quadrilaterals = 12650 - 630 - 35 = 11985

QUESTION: 15

In how many ways a committee consisting of 5 men and 3 women, can be chosen from 9 men and 12 women.

Solution:

Choose 5 men out of 9 men = 9C5 ways = 126 ways

Choose 3 women out of 12 women = 12C3 ways = 220 ways

The committee can be chosen in 27720 ways