Phrases Permutations and Combinations Previous Year Question Paper and Solution with PDF

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Phrases Previous Year Questions (PYQs)

Phrases Permutations And Combinations PYQ

Qus : 1

The Set of Intelligent Students in a class is

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Solution

Since, intelligency is not defined for students in a class i.e., Not a well defined collection.

Qus : 2

In a beauty contest, half the number of experts voted Mr. A and two thirds voted for Mr. B 10 voted for both and 6 did not for either. How may experts were there in all.

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Solution

Let the total number of experts be

N

E

is the set of experts who voted for miss

A

F

is the set of experts who voted for miss

B

.
Since

6

did not vote for either, n

(E \cup F) = N - 6

n (E) = \frac{N}{2}, n (F) = \frac{2}{3} N

and

n (E \cap F) = 10

.
So,

N - 6 = \frac{N}{2} + \frac{2}{3} N - 10

Solving the above equation gives

\frac{N}{6} = 4 \Rightarrow N = 24

Qus : 3

If all the words, with or without meaning, are written using the letters of the word QUEEN add are arranged as in English Dictionary, then the position of the word QUEEN is

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Solution

Letters of the word QUEEN are E,E,N,Q,U

Words beginning with E (4!) = 24

Words beginning with N (4!/2!)=12

Words beginning with QE (3!) = 6

Words beginning with QN (3!/2!)= 3

Total words = 24+12+6+9=45

QUEEN is the next word and has rank 46th.

Qus : 4

In a chess tournament, n men and 2 women players participated. Each player plays 2 games against every other player. Also, the total number of games played by the men among themselves exceeded by 66 the number of games that the men played against the women. Then the total number of players in the tournament is

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Solution

Qus : 5

There are 9 bottle labelled 1, 2, 3, ... , 9 and 9 boxes labelled 1, 2, 3,....9. The number of ways one can put these bottles in the boxes so that each box gets one bottle and exactly 5 bottles go in their corresponding numbered boxes is

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Solution

Total bottles and boxes: 9 each, labeled 1 to 9.

We are asked to count permutations of bottles such that exactly 5 bottles go into their own numbered boxes.

Step 1: Choose 5 positions to be fixed points (i.e., bottle number matches box number).

Number of ways = $\binom{9}{5}$

Step 2: Remaining 4 positions must be a derangement (no bottle goes into its matching box).

Let $D_4$ be the number of derangements of 4 items.

$D_4 = 9$

Step 3: Total ways = $\binom{9}{5} \times D_4 = 126 \times 9 = 1134$

✅ Final Answer: $\boxed{1134}$

Qus : 6

The value of

$\sum ^n_{r=1}\frac{{{{}^nP}}_r}{r!}$ is:

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Solution

Question: Find the value of:

$\sum_{r=1}^{n} \frac{nP_r}{r!}$

Solution:

We know: $nP_r = \frac{n!}{(n - r)!} \Rightarrow \frac{nP_r}{r!} = \frac{n!}{(n - r)! \cdot r!} = \binom{n}{r}$

Therefore,

$\sum_{r=1}^{n} \frac{nP_r}{r!} = \sum_{r=1}^{n} \binom{n}{r} = 2^n - 1$

Final Answer: $\boxed{2^n - 1}$

Qus : 7

Lines

$L_1, L_2, .., L_10$ are distinct among which the lines

$L_2, L_4, L_6, L_8, L_{10}$ are parallel to each other and the lines

$L_1, L_3, L_5, L_7, L_9$ pass through a given point C. The number of point of intersection of pairs of lines from the complete set

$L_1, L_2, L_3, ..., L_{10}$ is

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Solution

Total Number of Intersection Points

Given:

10 distinct lines: $L_1, L_2, \ldots, L_{10}$
$L_2, L_4, L_6, L_8, L_{10}$ : parallel (no intersections among them)
$L_1, L_3, L_5, L_7, L_9$ : concurrent at point $C$ (intersect at one point)

? Calculation:

$\text{Total line pairs: } \binom{10}{2} = 45$

$\text{Subtract parallel pairs: } \binom{5}{2} = 10 \Rightarrow 45 - 10 = 35$

$\text{Concurrent at one point: reduce } 10 \text{ pairs to 1 point} \Rightarrow 35 - 9 = \boxed{26}$

✅ Final Answer: $\boxed{26}$ unique points of intersection

Qus : 8

$\frac{n!}{2!(n-2)!}$ and

$\frac{n!}{4!(n-4)!}$ are in the ratio 2:1, then the value of n is

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Solution

Qus : 9

A polygon has 44 diagonals, the number of sides are

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Solution

Qus : 10

9 balls are to be placed in 9 boxes and 5 of the balls cannot fit into 3 small boxes. The number of ways of arranging one ball in each of the boxes is

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Solution

First off all select 5 boxes out 6 boxes in which 5 big ball can fit then arrange these ball in these 5 boxes and then put remaining 4 ball in any remaining box.

So Ans is [(6C5)5!](4!) = 6!4! = 17280

Qus : 11

A student council has 10 members. From this one President, one Vice-President, one Secretary, one Joint-Secretary and two Executive Committee members have to be elected. In how many ways this can be done?

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Solution

Qus : 12

How many natural numbers smaller than

can be formed using the digits 1 and 2 only?

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Solution

Qus : 13

A password consists of two alphabets from English followed by three numbers chosen from 0 to 3. If repetitions are allowed, the number of different passwords is

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Solution

Qus : 14

If n is an integer between 0 to 21, then find a value of n for which the value of

$n!(21-n)!$ is minimum

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Solution

Qus : 15

m distinct animals of a circus have to be placed in m cages, one is each cage. There are n small cages and p large animal (n < p < m). The large animals are so large that they do not fit in small cage. However, small animals can be put in any cage. The number of putting the animals into cage is

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Solution

Qus : 16

The number of ways in which 5 days can be chosen in each of the 12 months of a non-leap year, is:

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Solution

Qus : 17

Let A and B two sets containing four and two elements respectively. The number of subsets of the A × B, each having at least three elements is

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Solution

n(A) = 4

n(B) = 2

Then the number of subsets in A*B is 2⁸= 256

Qus : 18

, then the values of n and r are:

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Solution

Qus : 19

The number of ways to arrange the letters of the English alphabet, so that there are exactly 5 letters between a and b, is:

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Solution

Qus : 20

There are 50 questions in a paper. Find the number of ways in which a student can attempt one or more questions :

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Solution

Qus : 21

How many words can be formed starting with letter D taking all letters from the word DELHI so that the letters are not repeated:

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Solution

Qus : 22

Naresh has 10 friends, and he wants to invite 6 of them to a party. How many times will 3 particular friends never attend the party?

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Solution

^(10-3)C₆= ⁷C₆ = 7

Qus : 23

There is a young boy’s birthday party in which 3 friends have attended. The mother has arranged 10 games where a prize is awarded for a winning game. The prizes are identical. If each of the 4 children receives at least one prize, then how many distributions of prizes are possible?

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Solution

Qus : 24

$42 (^nP_2)=(^nP_4)$ then the value of n is

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Solution

Qus : 25

If n and r are integers such that 1 ≤ r ≤ n, then the value of n ^n-1C_r-1is

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Solution

Qus : 26

There are 8 students appearing in an examination of which 3 have to appear in Mathematics paper and the remaining 5 in different subjects. Then, the number of ways they can be made to sit in a row, if the candidates in Mathematics cannot sit next to each other is

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Solution

Qus : 27

The number of bit strings of length 10 that contain either five consecutive 0’s or five consecutive 1’s is

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Solution

Qus : 28

In an examination of nine papers, a candidate has to pass in more papers than the number of papers in which he fails in order to be successful. The number of ways in which he can be unsuccessful is

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Solution

Candidate Passing Criteria

Total Papers: 9

Condition for Success: Passes > Fails

So, candidate is unsuccessful when: Passes ≤ 4

Calculate ways:

$\text{Ways} = \sum_{x=0}^{4} \binom{9}{x} = \binom{9}{0} + \binom{9}{1} + \binom{9}{2} + \binom{9}{3} + \binom{9}{4}$ $= 1 + 9 + 36 + 84 + 126 = \boxed{256}$

✅ Final Answer: 256 ways

Qus : 29

How many even integers between 4000 and 7000 have four different digits?

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Solution

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Phrases Previous Year Questions (PYQs)

Phrases Permutations And Combinations PYQ

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Total bottles and boxes: 9 each, labeled 1 to 9.

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Total Number of Intersection Points

? Calculation:

✅ Final Answer: \boxed{26}\boxed{26} unique points of intersection

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✅ Final Answer: $\boxed{26}$ unique points of intersection