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Test: Permutation & Combination- 5


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15 Questions MCQ Test Quantitative Aptitude (Quant) | Test: Permutation & Combination- 5

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Test: Permutation & Combination- 5 - Question 1

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

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

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!

Test: Permutation & Combination- 5 - Question 2

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

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

14 numbers

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

Test: Permutation & Combination- 5 - Question 3

Detailed Solution for Test: Permutation & Combination- 5 - Question 3

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

Test: Permutation & Combination- 5 - Question 4

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

Detailed Solution for Test: Permutation & Combination- 5 - Question 4

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.

Test: Permutation & Combination- 5 - 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?

Detailed Solution for Test: Permutation & Combination- 5 - Question 5

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

Test: Permutation & Combination- 5 - Question 6

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

Detailed Solution for Test: Permutation & Combination- 5 - Question 6

In the ALLAHABAD :-
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.

Test: Permutation & Combination- 5 - 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.)

Detailed Solution for Test: Permutation & Combination- 5 - Question 7

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.

Test: Permutation & Combination- 5 - 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?

Detailed Solution for Test: Permutation & Combination- 5 - Question 8

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

Test: Permutation & Combination- 5 - 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?

Detailed Solution for Test: Permutation & Combination- 5 - Question 9

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). 

Test: Permutation & Combination- 5 - Question 10

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

Test: Permutation & Combination- 5 - 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)?

Detailed Solution for Test: Permutation & Combination- 5 - Question 11

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

Test: Permutation & Combination- 5 - 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?

Detailed Solution for Test: Permutation & Combination- 5 - Question 12

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

Test: Permutation & Combination- 5 - Question 13

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

Detailed Solution for Test: Permutation & Combination- 5 - Question 13

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 .

Test: Permutation & Combination- 5 - Question 14

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

Detailed Solution for Test: Permutation & Combination- 5 - Question 14

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

Test: Permutation & Combination- 5 - Question 15

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

Detailed Solution for Test: Permutation & Combination- 5 - Question 15

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

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