Permutations and Combinations jee mains questions | JEE Main Maths Permutations and Combinations Previous Year Question

Q 21 :

The lines $L_{1}, L_{2}, . . . L_{20}$ are distinct. For $n = 1, 2, 3, . . ., 10$ all the lines $L_{2 n - 1}$ are parallel to each other and all the lines $L_{2 n}$ pass through a given point P. The maximum number of points of intersection of pairs of lines from the set ${L_{1}, L_{2}, . . ., L_{20}}$ is equal to ______. [2024]

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(101)

Let $L_{1}, L_{3}, . . ., L_{19}$ are parallel to each other and $L_{2}, L_{4}, L_{6}, \dots, L_{20}$ are passing through a point P.

$∴$ Point of intersection of pairs of lines from the set ${L_{1}, L_{2}, \dots, L_{20}} = {}^{20}{C_{2} - {}^{10}{C_{2}} - {}^{10}{C_{2}}} + 1 = 101$

Q 22 :

In an examination of Mathematics paper, there are 20 questions of equal marks and the question paper is divided into three sections: A, B, and C. A student is required to attempt total 15 questions taking at least 4 questions from each section. If section A has 8 questions, section B has 6 questions and section C has 6 questions, then the total number of ways a student can select 15 questions is ______. [2024]

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(11376)

A B C Number of ways

5 6 4 ${}^{8}{C_{5}} \times {}^{6}{C_{6}} \times {}^{6}{C_{4}} = 56 \times 1 \times 15 = 840$

6 5 4 ${}^{8}{C_{6}} \times {}^{6}{C_{5}} \times {}^{6}{C_{4}} = 28 \times 6 \times 15 = 2520$

6 4 5 ${}^{8}{C_{6}} \times {}^{6}{C_{4}} \times {}^{6}{C_{5}} = 28 \times 15 \times 6 = 2520$

5 5 5 ${}^{8}{C_{5}} \times {}^{6}{C_{5}} \times {}^{6}{C_{5}} = 56 \times 6 \times 6 = 2016$

4 6 5 ${}^{8}{C_{4}} \times {}^{6}{C_{6}} \times {}^{6}{C_{5}} = 70 \times 1 \times 6 = 420$

4 5 6 ${}^{8}{C_{4}} \times {}^{6}{C_{5}} \times {}^{6}{C_{6}} = 70 \times 6 \times 1 = 420$

5 4 6 ${}^{8}{C_{5}} \times {}^{6}{C_{4}} \times {}^{6}{C_{6}} = 56 \times 15 \times 1 = 840$

7 4 4 ${}^{8}{C_{7}} \times {}^{6}{C_{4}} \times {}^{6}{C_{4}} = 8 \times 15 \times 15 = 1800$

$∴$ Required number of ways = 840 + 2520 + 2520 + 2016 + 420 + 420 + 840 = 11376

Q 23 :

The total number of words (with or without meaning) that can be formed out of the letters of the word 'DISTRIBUTION' taken four at a time, is equal to _______. [2024]

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(3734)

The word 'DISTRIBUTION' has total 12 letters and 9 diferent letters.

We select four different letters.

Case l : Three same letters and 1 different letter ${}^{1}{C_{1}} \times {}^{8}{C_{1}} \times \frac{4!}{3!}$

Case 2 :Two pairs are same and two pairs are different ${}^{2}{C_{1}} \times {}^{8}{C_{2}} \times \frac{4!}{2!}$

Case 3 : Two pair of letters are same ${}^{2}{C_{2}} \cdot {}^{4}{C_{2}}$

Case 4 : All different letters are select ${}^{9}{C_{4}} \times 4!$

Total letters selected.

$= {}^{1}{C_{1}} \times {}^{8}{C_{1}} \times \frac{4!}{3!} + {}^{2}{C_{1}} \times {}^{8}{C_{2}} \cdot \frac{4!}{2!} + {}^{2}{C_{2}} \cdot {}^{4}{C_{2}} + {}^{9}{C_{4}} \cdot 4!$

$= 1 \times \frac{8!}{7!} \times \frac{4!}{3!} + 2! \times \frac{8!}{6! 2!} \times \frac{4!}{2!} + \frac{1 \times 4!}{2! 2!} + \frac{9! \times 4!}{5! 4!}$

$= 32 + 672 + 6 + 3024 = 3734$

Q 24 :

There are 5 points $P_{1}, P_{2}, P_{3}, P_{4}, P_{5}$ on the side AB, excluding A and B, of a triangle ABC. Similarly, there are 6 points $P_{6}, P_{7}, \dots, P_{11}$ on the side BC and 7 points $P_{12}, P_{13}, \dots, P_{18}$ on the side CA of the triangle. The number of triangles, that can be formed using the points $P_{1}, P_{2}, \dots, P_{18}$ as vertices, is: [2024]

796
751
771
776

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2

3

4

(2)

Number of points on side AB = 5
Number of points on side BC = 6
Number of points on side AC = 7

Number of ways of selecting three points from side AB = ${}^{5}{C_{3}} $

Number of ways of selecting three points from side BC = ${}^{6}{C_{3}} $

Number of ways of selecting three points from side AC = ${}^{7}{C_{3}}$

Total number of triangles that can be formed using the points $P_{1}, P_{2}, \dots, P_{18} = {}^{18}{C_{3}} - ({}^{5}{C_{3} + {}^{6}{C_{3} + {}^{7}{C_{3}}}})$

$= {}^{18}{C_{3} - {}^{5}{C_{3}} - {}^{6}{C_{3}} - {}^{7}C_{3}} = 816 - 10 - 20 - 35 = 751$

Q 25 :

The number of triangles whose vertices are at the vertices of a regular octagon but none of whose sides is a side of the octagon is: [2024]

16
48
24
56

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2

3

4

(1)

Total number of triangles that can be formed by using 8 vertices of the octagon = ${}^{8}{C_{3}}$

Number of triangles having exactly one side common with the octagon = $8 \times 4$

Since, if we choose AB as the common side, then other vertices of the triangle will be either of G, F, E or D as we have exactly one common side.

So, we have $8 \times 4$ i.e., $32$ such triangles.

Now, let us find the number of triangles having two common sides.

In the octagon, we have 8 ways of choosing two consecutive sides, i.e., (AB, BC), (BC, CD), (CD, DE), (DE, EF), (EF, FG), (FG, GH), (GH, HA), (HA, AB)

$∴$ Number of triangles having 2 sides common with the octagon = 8

$∴$ Required number of triangles = ${}^{8}{C_{3}} - 32 - 8$

$= 56 - 32 - 8 = 16$

Q 26 :

If for some $m, n;$ ${}^{6}{C_{m}} + 2 ({}^{6}{C_{m + 1}}) +^{6} C_{m + 2} >^{8} C_{3}$ and ${}^{n - 1}{P_{3}} :^{n} P_{4} = 1 : 8,$ then ${}^{n}{P_{m + 1}} +^{n + 1} C_{m}$ is equal to [2024]

372
384
380
376

1

2

3

4

(1)

Given, ${}^{6}{C_{m}} + 2 ({}^{6}{C_{m + 1}}) +^{6} C_{m + 2} >^{8} C_{3}$

$\Rightarrow^{6} C_{m} +^{6} C_{m + 1} +^{6} C_{m + 1} +^{6} C_{m + 2} >^{8} C_{3}$

$\Rightarrow^{7} C_{m + 1} +^{7} C_{m + 2} >^{8} C_{3} \Rightarrow^{8} C_{m + 2} >^{8} C_{3} {or}^{8} C_{6 - m} >^{8} C_{3}$

$\Rightarrow m + 2 > 3$ i.e., $m > 1$

or $6 - m > 3$ i.e., $m < 3$ i.e., $1 < m < 3 \Rightarrow m = 2$

$∴$ ${}^{n}{P_{m + 1}} +^{n + 1} C_{m} =^{8} P_{3} +^{9} C_{2} = 336 + 36 = 372$

Q 27 :

The number of ways, in which the letters A, B, C, D, E can be placed in the 8 boxes of the figure below so that no row remains empty and at most one letter can be placed in a box is :

[2025]

5880
5760
840
960

1

2

3

4

(2)

We have 5 letters A, B, C, D, E.

The following table shows the numbers of ways to fill 5 boxes so that no row remains empty.

Row-I	Row-II	Row-III	Number of ways
3	1	1	${}^{3}C_{3} \cdot {}^{3}C_{1} \cdot {}^{2}C_{1} = 6$
2	2	1	${}^{3}C_{2} \cdot {}^{3}C_{2} \cdot {}^{2}C_{1} = 18$
1	3	1	${}^{3}C_{1} \cdot {}^{3}C_{3} \cdot {}^{2}C_{1} = 6$
2	1	2	${}^{3}C_{2} \cdot {}^{3}C_{1} \cdot {}^{2}C_{2} = 9$
1	2	2	${}^{3}C_{1} \cdot {}^{3}C_{2} \cdot {}^{2}C_{2} = 9$

Number of ways to fill boxes = 6 + 18 + 6 + 9 + 9 = 48

Now, these 5 letters can be arranged in 5! ways

$∴$ Total number of ways $48 \times 5! = 5760$ .

Q 28 :

Line $L_{1}$ of slope 2 and line $L_{2}$ of slop $\frac{1}{2}$ intersect at the origin O. In the first quadrant, $P_{1}, P_{2}, . . ., P_{12}$ are 12 points on line $L_{1}$ and $Q_{1}, Q_{2}, . . ., Q_{9}$ are 9 points on line $L_{2}$ . Then the total number of triangles, that can be formed having vertices at three of the 22 points O, $P_{1}, P_{2}, . . ., P_{12}, Q_{1}, Q_{2}, . . ., Q_{9}$ is : [2025]

1080
1026
1134
1188

1

2

3

4

(3)

Total number of triangles are

$= {}^{9}C_{1} {}^{12}C_{2} + {}^{9}C_{2} {}^{12}C_{1} + {}^{1}C_{1} \cdot {}^{9}C_{1} {}^{12}C_{1}$

= 594 + 432 + 108 = 1134

Q 29 :

From a group of 7 batsmen and 6 bowlers, 10 players are to be chosen for a team, which should include atleast 4 batsmen and atleast 4 bowlers. One batsmen and one bowler who are captain and vice-captain respectively of the team should be included. Then the total number of ways such a selection can be made, is [2025]

145
155
165
135

1

2

3

4

(2)

To select 10 players including atleast 4 batsmen and 4 bowlers.

Captain and vice-captain already selected

$∴$ Total number of ways = ${}^{6}C_{5} \times {}^{5}C_{3} + {}^{6}C_{4} \times {}^{5}C_{4} + {}^{6}C_{3} \times {}^{5}C_{5}$

= $6 \times 10 + 15 \times 5 + 20 \times 1$

= 60 + 75 + 20 = 155

Q 30 :

There are 12 points in a plane, no three of which are in the same straight line, except 5 points which are collinear. Then the total number of triangles that can be formed with the vertices at any three of these 12 points is [2025]

200
210
230
220

1

2

3

4

(2)

The number of ways to choose 3 points from 12 = ${}^{12}C_{3}$

= $\frac{12 \times 11 \times 10}{3 \times 2 \times 1}$ = 220

Number of ways to choose 3 points from 5 collinear points = ${}^{5}C_{3}$ = 10

Since these 5 points are collinear, so they can not form any triangles.

$∴$ Number of triangles that can be formed = 220 –10 = 210

Topic Question Set

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Q 23 :

The total number of words (with or without meaning) that can be formed out of the letters of the word 'DISTRIBUTION' taken four at a time, is equal to _______. [2024]

0%

Q 25 :

The number of triangles whose vertices are at the vertices of a regular octagon but none of whose sides is a side of the octagon is: [2024]

Q 26 :

If for some $m, n;$ ${}^{6}{C_{m}} + 2 ({}^{6}{C_{m + 1}}) +^{6} C_{m + 2} >^{8} C_{3}$ and ${}^{n - 1}{P_{3}} :^{n} P_{4} = 1 : 8,$ then ${}^{n}{P_{m + 1}} +^{n + 1} C_{m}$ is equal to [2024]

Q 27 :

The number of ways, in which the letters A, B, C, D, E can be placed in the 8 boxes of the figure below so that no row remains empty and at most one letter can be placed in a box is :

[2025]

Q 30 :

There are 12 points in a plane, no three of which are in the same straight line, except 5 points which are collinear. Then the total number of triangles that can be formed with the vertices at any three of these 12 points is [2025]

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