There
are 3 containers and 5 items. In how many ways can these items be placed in the
containers?
3+3 = 6 ways
TypeII(B) No barring on number of items of each container:
3+6+3+6+3 = 21 ways
TypeIII: Similar Containers – Different Items:
Examples:
This
is a conceptual one and we can see four
different cases here:
TypeI: Similar Containers – Similar Items:
X

X

X

5

0

0

4

1

0

3

1

1

3

2

0

2

2

1

5
ways
TypeII: Different Containers – Similar Items:
TypeII(A) Each
container having at least one item:
X

Y

Z


3

1

1

3!/2! = 3 ways

2

2

1

3!/2! = 3 ways

If the number of containers is r and the number of items is
n, and if each container should contain at least one item,
then the number of ways = (n1)C(r1)
= (51)C(31)
= 4C2 = 6TypeII(B) No barring on number of items of each container:
X

Y

Z


5

0

0

3!/2! = 3
ways

4

1

0

3! = 6
ways

3

1

1

3!/2! = 3
ways

3

2

0

3! = 6
ways

2

2

1

3!/2! = 3
ways

3+6+3+6+3 = 21 ways
If
the number of containers is r and the number of items is n, the formula here is (n+r1)C(r1)
=
(5+31)C(31) = 7C2 = 21
TypeIII: Similar Containers – Different Items:
X

X

X


5

0

0

1 way

4

1

0

5!/4! = 5
ways

3

1

1

5!/3! = 20
ways

3

2

0

5!/3!2! =
10 ways

2

2

1

5!/2!2! =
30 ways

1+5+20+10+30
= 66 ways
TypeIV: Different Containers – Different Items:
This
is combination of types II and III.
X

Y

Z


5

0

0

3*1
= 3 ways

4

1

0

6*5 = 30
ways

3

1

1

3*20 = 60
ways

3

2

0

6*10 = 60
ways

2

2

1

3*30 = 90
ways

3+30+60+60+90
= 243 ways
If
the number of containers is r and the number of items is n, the formula here
is
r^{n }=
3^{5} =
243
Most
of the problems can be categorized into one of these four types.
Examples:
Q1)There are 5 types of Icecreams in a shop. In how many different ways can Rahul buy 6 Icecreams?
Solution:
Here selecting 6 Icecreams (may be of any variety) out of 5 different varieties is the only criteria and no further arrangements of the selected Icecreams is required. Hence, this can be categorized as TypeII(B) problem, where number of different varieties of Icecreams = r = 5 and number of Icecreams to buy = n = 6
Here selecting 6 Icecreams (may be of any variety) out of 5 different varieties is the only criteria and no further arrangements of the selected Icecreams is required. Hence, this can be categorized as TypeII(B) problem, where number of different varieties of Icecreams = r = 5 and number of Icecreams to buy = n = 6
=>
(n+r1)C(r1) = 10C4
ways
Q2) A new flag has to be designed with six vertical stripes using some or all of the colours yellow, green, blue and red. Then the number of ways this can be done such that no two adjacent stripes have the same colour is: (CAT2004)
Note that had there been no condition that "no two adjacent stripes have the same colour", the answer would have been r^{n} = 4^{6}
^{}
Concept on number of integral solutions:
Q4)Find the total number of natural numbers having sum of digits five and number of digits at most five?
Q2) A new flag has to be designed with six vertical stripes using some or all of the colours yellow, green, blue and red. Then the number of ways this can be done such that no two adjacent stripes have the same colour is: (CAT2004)
(1)12*81 (2)16*192 (3)20*125 (4)24*216
Solution:
Here
selecting 6 colours out of 4 different varieties and also arranging them is required
(such that no two adjacent stripes have the same colour).
Hence, this
can be categorized as TypeIV problem.
First
stripe can have 4 choices, second one can have remaining 3 choices, again third
one can have 3 choices, similarly, fourth, fifth and sixth ones can have three
choices each.
∴Total number of ways =
4*3*3*3*3*3 = 12*81
Answer (1)Note that had there been no condition that "no two adjacent stripes have the same colour", the answer would have been r^{n} = 4^{6}
^{}
Concept on number of integral solutions:
Q3) To find the number of integral solutions of the
equation x_{1}+x_{2}+x_{3}+....+x_{r} = n:
This
is similar to TypeII model with ‘r’ number of baskets and ‘n’ number of items.
For finding number of nonnegative integral solutions:
The solution may contain zeroes also, hence it is similar to
TypeII(B) model
=> Number of solutions = (n+r1)C(r1)
For finding number of positive integral solutions:
The solution should not contain zeroes, hence it is similar
to TypeII(A) model
=>
Number of solutions = (n1)C(r1)Q4)Find the total number of natural numbers having sum of digits five and number of digits at most five?
Solution:
Here, the number of digits is at most five. Hence it may be 1
or 2 or 3 or 4 or 5.
_ _ _ _ _
If the left most digit is filled with any nonzero number,
then this set becomes a five digit number.
Leaving the left most digit vacant, if the secondleft digit
is filled, this set becomes a 4digit number,
and so on and so forth.. Finally, leaving
four left most digits vacant, if the unit digit is filled with any nonzero
number, than the set becomes a singledigit number. So we can consider 5 digits at a time and depending on the position of zeroes
from the left, different sized numbers will be formed.
As we need to get the sum of digits also
to be 5, this is of model TypeIIB with the number of containers being 5 and
number of items also being 5.
(5+51)C(51) = 9C4 = (9*8*7*6)/(4*3*2*1)
= 126
Q5)In
how many ways five different coins can be distributed among three persons where
every one get at least one coin?
Solution:
Here, the coins as well as the persons being different, this comes
to be TypeIV.
A

B

C

Ways

3

1

1

3*(5!)/(3!) =
60 ways

2

2

1

3*(5!)/(2!)(2!)
= 90 ways

60+90 = 150 ways
Explanation:
First, these five different coins can be divided into 3,1,1
set in 5!/3! ways
Next this (3,1,1) set can be distributed among three in the following
3 ways:
(3,1,1); (1,3,1); (1,1,3);
Combining, we get, 3*5!/3! = 60
Similarly (2,1,1) set has 3*5!/(2!*2!) = 90 ways.
In total, 60+90 = 150 ways.
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