Permutation & combination
Questions
Q1.
The total number of hand shakes is 66, when everybody in a group shakehands with everybody once . Find the number of persons
in a group .
(a) 12
(b) 10
(c) 6
(d) 11
Solution.
Formula
nC
2 = No . of shakehands
nC
2 = 66
n(n-1)/2
= 66 ⇒ n = 12
Answer = 12
Solution
Q2.
The total number of hand shakes is 45, when everybody in a meeting shakehands with everybody . Find the number of persons
in a group .
(a) 8
(b)10
(c) 5
(d) 12
No . of shakehands =
nC
2
n(n-1)/2
= 45 ⇒ n = 10
Answer = 10
Q3.
In a room there are 6 lamp, have option switch off and on . Each can be worked independently . Find the number of ways in which
room can be illuminated .
(a) 60
(b) 62
(c) 56
(d) 63
Solution.
Each lamp has two option either switched on or off , So required number = 26 - 1 = 63
Note : There
is one way when all lamps will switched off and room will not illuminated .
Answer = 63
Solution
Q4.
Find the number of ways in which a selection of 4 letters can be made from the word 'MATHEMATICS' .
(a) 100
(b) 136
(c) 163
(d) 120
Solution.
There are 11 letters in the word of which A, M, T are repeated twice.
Thus we have 11 letters of 8 differents kinds (A,A) , (M, M) , (T, T) , H, E ,
I, C, S .
The group of four selected letter may take any of the following forms :
(i) Two alike and other
two alike
(ii) Two alike and other two different
(iii) All four different
In case (i) , the number
of ways =
3C
2 = 3
In case (ii) , the number of ways =
3C
1 x
7C
2 = 3 x 21 = 63 .
In case (iii) , the number of ways =
8C
4 =
8x7x6x5/1x2x3x4
= 70
Hence, the required number of ways = 3 + 63 + 70 = 136 ways
Answer = 136 ways
Solution
Q5.
Find the number of ways in which a selection of 4 letters can be made from the word 'Examination'.
(a) 123
(b) 100
(c) 155
(d) 136
Solution.
There are 11 letters in the word of which A,I,N are repeated twice.
Thus we have 11 letters of 8 differents kinds (A,A) , (I,I) , (N,N) , E ,
X , M , T , O .
The group of four selected letter may take any of the following forms :
(i) Two alike and other
two alike
(ii) Two alike and other two different
(iii) All four different
In case (i) , the number
of ways =
3C
2 = 3
In case (ii) , the number of ways =
3C
1 x
7C
2 = 3 x 21 = 63 .
In case (iii) , the number of ways =
8C
4 =
8x7x6x5/1x2x3x4
= 70
Hence, the required number of ways = 3 + 63 + 70 = 136 ways
Answer = 136 ways
Solution
