Permutation and Combination

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41.

On the occasion of a certain meeting each member gave shakehand to the remaining members. If the total shakehands were 28, how many members were present for the meeting?

View Answer

Shortcut:
In a function every man shakes hand with every other man. If there was a total of H handshakes in the function, the number of men 'n' who were present in the function can be calculated by the following equation.

n(n − 1) / 2

= H.
Here, n = ?, H = 28
Using the value in the shortcut, we get:

n(n − 1) / 2

= 28
n(n − 1) = 2 x 28
n(n − 1) = 56
n(n − 1) = 8 x 7
n = 8
Hence, 8 members were present for the meeting.

42.

There are 5 members in a delegation which is to be sent abroad. The total number of members is 10. In how many ways can the selection be made so that a particular member is always (i) included (ii) excluded?

(i): 132; (ii): 138

(i): 126; (ii): 126

(i): 120; (ii): 130

(i): 125; (ii): 115

View Answer

Shortcut:
There are 'm' members in a delegation which is to be sent abroad. The total number of members is 'z'. The number of ways in which the selection can be made so that a particular 'n' members are always
(i) included is given by ^{z − n}C_{m − n}
(ii) excluded, is given by ^{z − n}C_m
Here, m = 5, z = 10, n = 1
Using the value in the shortcut, we get:
(i) included is given by ^{10 − 1}C_{5 − 1} = ⁹C₄
=

9! / 4!(9 − 4)!

9! / 4! x 5!

9 x 8 x 7 x 6 x 5! / 4 x 3 x 2 x 1 x 5!

= 9 x 7 x 2 = 126
(ii) excluded, is given by ^{10 − 1}C₅ = ⁹C₅
=

9! / 5!(9 − 5)!

9! / 5! x 4!

9 x 8 x 7 x 6 x 5! / 5! x 4 x 3 x 2 x 1

= 9 x 7 x 6 = 126

43.

There are 10 points in a plane out of which 4 are collinear. Find the number of triangles formed by the points as vertices.

120

116

118

110

View Answer

Shortcut:
There are 'n' points in a plane out of which 'm' points are collinear. The number of triangles formed by the points as vertices are given by (ⁿC₃ − ^mC₃)
Here, n = 10, m = 4
Using the value in the shortcut, we get:
¹⁰C₃ − ⁴C₃ =

10! / 3!(10 − 3)!

−

4! / 3! x (4 − 3)!

10! / 3! x 7!

−

4! / 3! x 1!

10 x 9 x 8 x 7! / 3 x 2 x 1 x 7!

−

4 x 3! / 3! x 1!

= 5 x 3 x 8 − 4
= 120 − 4 = 116

44.

There are 10 points in a plane out of which 4 are collinear. Find the number of straight lines formed by joining them.

View Answer

Shortcut:
There are 'n' points in a plane out of which 'm' points are collinear. The number of straight lines formed by the joining them are given by (ⁿC₂ − ^mC₂ + 1)
Here, n = 10, m = 4
Using the value in the shortcut, we get:
¹⁰C₂ − ⁴C₂ + 1 =

10! / 2!(10 − 2)!

−

4! / 2! x (4 − 2)!

+ 1
=

10! / 2! x 8!

−

4! / 2! x 2!

+ 1
=

10 x 9 x 8! / 2! x 8!

−

4 x 3 x 2 x 1 / 2 x 1 x 2 x 1

+ 1
= 5 x 9 − 3 x 2 x 1 + 1
= 45 − 6 + 1
= 45 − 5 = 40

45.

On a new year day every student of a class sends a card to every other student. The postman delivers 600 cards. How many students are there in the class.

View Answer

Shortcut:
On a new year day every student of a class sends a card to every other student. If the postman delivers 'Z' cards, then the number of students 'n' in the class can be calculated by the following equation n(n − 1) = Z
Here, n = ?, Z = 600
Using the value in the shortcut, we get:
n(n − 1) = 600
n(n − 1) = 25 x 24
n = 25

46.

In an examination a minimum is to secured in each of 5 subjects for a pass. In how many ways can a student fail?

31 ways

35 ways

33 ways

25 ways

View Answer

Shortcut:
If in an exam a minimum is to be secured in each of 'n' subjects for a pass, then the number of ways a student can fail is 2ⁿ − 1.
Here, n = 5
Using the value in the shortcut, we get:
= (2)⁵ − 1
= 32 − 1 = 31
Hence, a student can fail in 31 ways.

47.

There are 6 questions in a question paper. In how many ways can a student solve one or more questions?

View Answer

Shortcut:
If in an exam a minimum is to be secured in each of 'n' subjects for a pass, then the number of ways a student can fail is 2ⁿ − 1.
Here, n = 6
Using the value in the shortcut, we get:
= (2)⁶ − 1
= 64 − 1 = 63
Hence, a student can solve one or more questions in 63 ways.

48.

From 4 officers and 8 jawans in how many ways can 6 be chosen to include exactly one officer?

225

220

224

230

View Answer

Shortcut:
From 'a' persons of a group X and 'b' persons from group Y, the number of ways in which 'n' persons can be chosen to include exactly 'p' persons of group X and the rest of group Y is given by (^aC_p x ^bC_{n − p}) ways
Here, a = 4, b = 8, n = 6, p = 1
Using the value in the shortcut, we get:
Required ways = (⁴C₁ x ⁸C_{6 − 1})
= (⁴C₁ x ⁸C₅)
=

4! / 1!(4 − 1)!

8! / 5! x (8 − 5)!

4! / 1! x 3!

8! / 5! x 3!

4 x 8 x 7 x 6 x 5! / 5! x 3 x 2 x 1

= 4 x 8 x 7 = 224
Hence, we can choose in 224 ways.

49.

There are 4 oranges, 5 apples and 6 mangoes in a fruit basket. In how many ways can a person make a selection of fruits from among the fruits in the basket?

215

209

205

213

View Answer

Shortcut:
In a basket there are certain number of fruits. Out of which, there are 'a' oranges, 'b' apples, 'c' mangoes and the remaining 'z' are different kinds. Then the number of ways a person can make a selection of fruits from among the fruits in the basket are given by [(a + 1)(b + 1)(c + 1) x 2^z − 1] ways
Here, a = 4, b = 5, c = 6, z = 0
Using the value in the shortcut, we get:
Required ways = [(4 + 1)(5 + 1)(6 + 1) x (2⁰ − 1)] ways
= 5 x 6 x 7 x 1 − 1
= 210 − 1 = 209
Hence, we can choose in 209 ways.

50.

In how many ways 12 different things can be divided equally among 3 persons?

View Answer

Shortcut:
The number of ways in which (a x b) different things can be divided equally among 'a' persons are given by

(ab)! / (b!)^a

Here, a x b = 12 = 3 x 4
or, a = 3, b = 4
Using the value in the shortcut, we get:
Required ways =

(3 x 4)! / (4!)³

12! / (4!)³