Combinations
Long Answers Questions Type-I
1. A
student has to answer 10 questions, choosing at least 4 from each of part A and
B. If there are 6 questions in part A and 7 in part B. In how many ways can the
student choose 10 questions?
Sol.
Combination
from A and from B:: 
; 
; 
Number of ways to get
pattern = 6C4 
7C6
=
= 157
=
105
Number of ways to get pattern = 6C5 7C5
=
= 621
= 126
Number of ways to get
pattern = 6C67C4
=
= 35
Hence,
total number of ways
=
105+126+35
=
266
2.
From a class of 15 students, 10 are to chosen for a picnic.
There are two students who decide that either both will join or none of them
will join. In how many ways can the picnic be organised?
Sol.
Case 1:
Both
join
In case both the
students decide to join,
= 13C8
2C2 ways
= 13C8
= 
=1287
Case 2: In
case none of them join, it will be 13C10
Ways 13C10
= 
= 286
Total number of cases are
= 13C8
+ 13C10
=
1287 +286
= 1573
3.
How many different products can be obtained by multiplying two or more of the
numbers 2, 5, 6, 7, 9?
Sol.
The given numbers are 2, 5, 6, 7, and 9.
The numbers of
different products when 2 or more is taking = the number of ways of taking
product of 2 numbers + number of ways of taking product of 3 numbers + number
of ways of taking product of 4 numbers + number of ways of taking 5 together
= 5C2 +
5C3 + 5C4 + 5C5
= 
+ 
+
+
=
+ 
+ 
+ 1
=26
4.
Determine the number of 5 cards combinations out of pack of 52 cards if at least
3 out of 5 cards are ace cards?
Sol.
There are 4 ace card in pack of 52 cards, therefore we can
choose maximum 4 ace cards.
Case 1: 3 cards of ace and 2 cards out of
remaining 4 cards
i.e. 5 cards combinations
out of pack of 52 cards
= 4C3 48C2
= 
+ 
=
=
4512
Case 2: 4 cards of ace and 1 cards out of 52
remaining cards
i.e. 5 cards combinations
out of 52 cards
= 4C4 48C1
= 
+ 
=
=
48
Hence, total number
of combinations are
= 4512 +
48
= 4560
5.
Find the number of all possible arrangement of the letter of the word
“MATHEMATICS” taken four form at a time?
Sol.
The word MATHEMATICS consists of 11 letters:
(M, M), (A, A), (T, T), H, E, I, C, S
Case 1: In this case 2 similar and 2 similar
letters are selected, number of arrangements
=
3C2 
=
18
Case 2: In this case 2 similar and 2 different
letters are selected, number of arrangements
=
3C1 7C2 
=
756
Case 3: In this case all 4 letters selected are different,
number of arrangements
= 8C4 4!
=
1680
Therefore, total number of arrangements
=
18 + 756 + 1680
=
2454
6.
Three married couples are to be seated in a row having six seats in a cinema hall.
If spouses are to be seated to each other, in how many ways can they be seated?
Find also the number of ways of their seating if all the ladies sit together.
Sol.
(i) Three couples can be seated in a
row in 3P3 =3! Ways.
Now, in each couple,
the spouse can be arranged in 2P2 =2! Ways.
Thus for three
couples, number of arrangement = 2! 2! 2!
Total number of ways
in which spouses are seated next to each other
=
3! 2! 2! 2!
= 6 2 2 2
= 48
Ways.
(ii) Now, if the three ladies are to be seated together, then we
regard 3 ladies as one block. Therefore, there are now 4 people i.e., 3 gents
and 1 block of 3 ladies
These 4 people can be
arranged in 4P4 =4!
=
24 Ways
But three ladies can
interchange their position in 3! = 6 ways
Total number of
arrangements in which 3 ladies sit together
=
24 6
= 144.
Long Answers Questions Type-II
1.
The set S = {1, 2, 3… 12} is to be partitioned into three sets A, B, and C of
equal sizes.
A
BC = S, A
B = BC =CA =
. Find the number of ways to partitions.
Sol.
There are 12 elements.
Three disjoint sets
A, B and C of same size are to be formed
Each set has 4
elements.
Number of ways of
selecting any 4 elements from 12 elements for set A = 12C4
Number of ways of
selecting any 4 elements from 8 elements for set B = 8C4
Number of ways of
selecting any 4 elements from 4 elements for set A = 4C4
Total number of
partition = 12C4 X 8C4 X 4C4.
2.
Find the value of 50C4 + 
56-rC3
Sol.
Given expression is
50C4
+ 56-rC3
= 50C4 + 55C3 + 54C3
+ 53C3 + 52C3 + 51C3
+ 50C
3
Writing
the terms in reverse order, we get
=
(50C4 + 50C3) + 51C3+
52C3+ 53C3+ 54C3+
55C3
= 51C4+ 51C3+
52C3+ 53C3+ 54C3+
55C3
[
nCr+
nCr-1 = n+1Cr]
=
(51C4+ 51C3) + 52C3+
53C3+ 54C3+ 55C3
= (52C4+ 52C3)
+ 53C3+ 54C3+ 55C3
= (53C4+ 53C3)
+ 54C3+ 55C3
= (54C4+ 54C3)
+ 55C3
=
55C4+ 55C3
= 56C4
3.
From 6 different novels and 3 different dictionaries, 4 novels and a dictionaries
is to be selected and arranged in a row on the shelf so that the dictionary is
always in the middle. Then find the number of such arrangements.
Sol.
The number of ways in which 4 novels can
be selected from 6 different novels = 6C4 =15
The number of ways in
which 1 dictionary can be selected from 3 different dictionaries = 3Cr =3
Now, it’s given that dictionary
is always in the middle.
Hence, the
arrangement look like M M D M M
Dictionary is always
in the middle as novels are different, hence they can be arranged in 4! Ways.
Hence, total number
of such arrangement
= 15 × 3 × 4 × 3 × 2 × 1
= 1080