Test: Permutation And Combination- 5


15 Questions MCQ Test UPSC Prelims Paper 2 CSAT - Quant, Verbal & Decision Making | Test: Permutation And Combination- 5


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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:

Correct Answer:B

14 numbers

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

QUESTION: 3

Solution:
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:
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:

Total number of 6 digit numbers having 3 odd and 3 even digits (including zero in the left most place) = 53 x 53.
From this subtract the number of 5 digit numbers with 2 even digits and 3 odd digits (to take care of the extra counting due to zero)

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:

The possible cases for counting are: Number of numbers when the units digit is nine + the number of numbers when neither the units digit nor the left most is nine + number of numbers when the left most digit is nine.

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 condition is that we have to count the number of natural numbers not more than 4300.
The total possible numbers with the given digits = 5x5 x5 x5 = 625 - 1 = 624.
Subtract form this the number of natural number greater than 4300 which can be formed from the given digits =1x 2x 5x5 -1 =49 .
Hence, the required number of numbers = 624 - 49 = 575.

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:

The number of numbers formed would be given by 5 x 4 x 3 (given that the first digit can be filled in 5 ways, the second in 4 ways and the third in 3 ways - MNP rule).

QUESTION: 12

In the above question, what will be the number of ways of selecting the committee with at least 3 women such that at least one woman holds the post of either a president or a vice-president?

Solution:

For a straight line we just need to select 2 points out of the 8 points available. 8C2 would be the number of ways of doing this.

QUESTION: 13

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

Solution:

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