GRAPH THEORY AND APPLICATIONS - EXPONENTIAL GENERATING FUNCTION

1. Exponential Generating function (EGF):
1) Defn:-
For the real sequence a0,a1,a2,…
E(x)= a0+𝑎1
𝑥
2!
+ 𝑎2
𝑥2
2!
+ 𝑎3
𝑥3
3!
+ ⋯ = ∑ 𝑎 𝑛
𝑥 𝑛
𝑛!
∞
𝑛=0
Is called the exponential generating function.
2)Examples
Sequence E(x)
1,1,1,1,… 1+
𝑥
1!
+
𝑥2
2!
+ ⋯ = 𝑒 𝑥
1,-1,1,-1,… 1-
𝑥
1!
+
𝑥2
2!
−
𝑥3
3!
… = 𝑒−𝑥
1,2,22
,23
,… 1+
2𝑥
1!
+ 22 𝑥2
2!
+ ⋯ = 𝑒2𝑥
1,-a,𝑎2
,−𝑎3
, … 1-
𝑎𝑥
1!
+ 𝑎2 𝑥2
2!
− 𝑎3 𝑥3
3!
… = 𝑒2𝑥
1,𝑎2
, 𝑎4
, 𝑎6
… 1+𝑎2 𝑥
1!
+ (𝑎2
)2 𝑥2
2!
+ ⋯ = 𝑒 𝑎2 𝑥
a,𝑎3,
, 𝑎5
, 𝑎7
, … a+𝑎3 𝑥
1!
+ 𝑎5 𝑥2
2!
+ ⋯ = 𝑎(1+𝑎2 𝑥
1!
+
(𝑎2
)2 𝑥2
2!
+ ⋯)=a𝑒 𝑎2 𝑥
0,1,2(2),3(22),… 0+1.
𝑥
1!
+2(2)
𝑥2
2!
+3(22
)
𝑥3
3!
+…=
X[1+
2𝑥
1!
+
(2𝑥)2
2!
…] =
x𝑒2𝑥
3)𝑒 𝑥
+ 𝑒−𝑥
=2(1+
𝑥2
2!
+
𝑥4
4!
+…)
𝑒 𝑥
− 𝑒−𝑥
= 2(𝑥 +
𝑥3
3!
+
𝑥5
5!
+ ⋯
4) In how many ways can your of the letters in ENGINE be arranged?
Given:
Word = ENGINE

2. Aim: To find the no. of ways can four of the letters E, N, G, I, in the given
word is arranged.
Solution:
S1)
Letter Sequence f(x)
E 0,1,2 1+
𝑥
1!
+
𝑥2
2!
N 0,1,2 1+
𝑥
1!
+
𝑥2
2!
G 0,1 1+
𝑥
1!
I 0,1 1+
𝑥
1!
S2) E(x) =(1+
𝑥
1!
+
𝑥2
2!
) (1+
𝑥
1!
+
𝑥2
2!
)( 1+
𝑥
1!
)( 1+
𝑥
1!
)
S3) Method I
The sequence ans=co.eff of
𝑥4
4!
in E(x)
E(x) =(1+
𝑥
1!
+
𝑥2
2!
𝑥 +
𝑥2
2!
+
𝑥3
2!
+
𝑥2
2!
+
𝑥3
2!
+
𝑥4
2!2!
)
E(x) =(1+2x+3𝑥2
+𝑥3
+
𝑥4
4
)(1+2x+2𝑥2
)
E(x) =Co.eff of x4 [2+2+
1
4
]
=co.eff of x4[4+
1
4
]
E(x) =co.eff of
𝑥4
4!
[(
17
4
)4!]
∴ The sequence ans=co.eff of
𝑥4
4!
In E(x)
The sequence ans=
17
4
. 4!=
17
4
× 4 × 3 × 2 × 1=102
S4)Method II P=
𝑚!
𝑛1!𝑛2!𝑛3!…

3. Possible selection of four letters N0. Of Arrangement
E E G N 4!
2!1!1!…
=12
E E N N 4!
2!2!…
=6
E E I N 4!
2!1!1!…
=12
E E G I 4!
2!1!1!…
=12
E G N N 4!
2!1!1!…
=12
E I N N 4!
2!1!1!…
=12
G I N N 4!
2!1!1!…
=12
E I G N 4!
2!1!1!…
=24
102
Hence 102 ways 4 letters word can be arranged from the given word ENGINE
5) A ship carries 48 flags, 12 each of the colours Red, White, Blue, and Black.
Twelve of these flags are placed on a vertical pole in order to communicate a
signal to other ships.
a) How many of these signals use an even number of blue flags and an odd
number of black flags?
b) How many of the signals have atleast three white flags or no white flags or
no white flags at all?
Given:
Total no. of flags =48
No. of colours=4
No. of flags in a signal =12
Aim: To find no. of ways the 12 signal flags arrange such that
i) Even no. of blue and off no. of blacks
ii) Atleast 3 white or no white flag
Solution:
i)

4. Colour Sequence f(x)
Red 0,1,2,…,12 1+
𝑥
1!
+
𝑥2
2!
+…+
𝑥12
12!
Blue
(even)
0,2,4,6,8,10,12 1+
𝑥2
2!
+
𝑥4
4!
+…+
𝑥12
12!
Black
(odd)
1,3,5,7,9,11 𝑥
1!
+
𝑥3
3!
+…+
𝑥11
11!
White 0,1,2,3,4,…12 1+
𝑥
1!
+
𝑥2
2!
+…+
𝑥12
12!
⇒E(x)=𝜋f(x)=(
𝑥0
0!
+
𝑥2
2!
+
𝑥4
4!
+ ⋯
𝑥12
12!
)(
𝑥
1!
+
𝑥3
3!
+…+
𝑥11
11!
)(1+
𝑥
1!
+
𝑥2
2!
+…+
𝑥12
12!
)(1+
𝑥
1!
+
𝑥2
2!
+…+
𝑥12
12!
)
Suppose n≥ 1,
E(x)=( 1+
𝑥2
2!
+
𝑥4
4!
+ ⋯)(
𝑥
1!
+
𝑥3
3!
+
𝑥5
5!
+ ⋯)(1+
𝑥
1!
+
𝑥2
2!
+…)2
=(
𝑒 𝑥 +𝑒 −𝑥
2
)(
𝑒 𝑥−𝑒−𝑥
2
)(𝑒 𝑥
)2
E(x)=
1
4
𝑒2𝑥
(𝑒2𝑥
-𝑒−2𝑥
)=
1
4
(𝑒4𝑥
− 1)
ie, E(x)=
1
4
[∑
(4𝑥) 𝑛
𝑛!
∞
𝑛=0 -1]
=
1
4
[1 + ∑
(4𝑥) 𝑛
𝑛!
∞
𝑛=1 − 1]
E(x)=
1
4
∑
4 𝑛 𝑥 𝑛
𝑛!
∞
𝑛=1
But the seq ans=co.eff of
𝑥12
12!
In E(x)
=
1
4
412
=411
iii) In this case,
Colour Sequence f(x)
Red 0,1,2,…,12 1+
𝑥
1!
+
𝑥2
2!
+…+
𝑥12
12!
Blue 0,1,2,…,12 1+
𝑥
1!
+
𝑥2
2!
+…+
𝑥12
12!
Black 0,1,2,…,12 1+
𝑥
1!
+
𝑥2
2!
+…+
𝑥12
12!
White 0,3,4,5,5,6,…,12 1+
𝑥3
3!
+
𝑥4
4!
+…+
𝑥12
12!

5. E(x)= ∏ 𝑓1 (𝑥)4
𝑖=1 =(1 +
𝑥
1!
+
𝑥2
2!
+ ⋯+
𝑥12
12!
)3
(1+
𝑥3
3!
+
𝑥4
4!
+…+
𝑥12
12!
)
Supposen≥ 1
E(x)=𝑒3𝑥
(𝑒 𝑥
− 𝑥 −
𝑥2
2!
)=𝑒4𝑥
− 𝑥𝑒3𝑥
−
𝑥2
2
𝑒3𝑥
E(x)=∑
4 𝑛 𝑥 𝑛
𝑛!
∞
𝑛=0 − 𝑥 ∑
3 𝑛 𝑥 𝑛
𝑛!
∞
𝑛=0 -
𝑥2
2
∑
3 𝑛 𝑥 𝑛
𝑛!
∞
𝑛=0
The sequence ans=Sum of the co-eff of
𝑥12
12!
In each term
=412
− 12(311)-
1
2
(12)(11)(310)
=10,754,218
6) A company river 11 new employees, each of whom is to be
assigned to one of four subdivision. Each subdivision will get atleast
one new employee. In how many ways can these assignment be made?
Given:
No. of subdivision =4 (say A,B,C,D)
No. of new employees=11
Aim:
To find no. of ways that each subdivision get atleast one new
employee.
Solution:
Subdivision Sequence f(x)
A 1,2,3,…11 𝑥
1!
+
𝑥2
2!
+…+
𝑥11
11!
B 1,2,3,…11 𝑥
1!
+
𝑥2
2!
+…+
𝑥11
11!
C 1,2,3,…11 𝑥
1!
+
𝑥2
2!
+…+
𝑥11
11!
D 1,2,3,…11 𝑥
1!
+
𝑥2
2!
+…+
𝑥11
11!
E(x)=(
𝑥
1!
+
𝑥2
2!
+…+
𝑥11
11!
)4 =(𝑒 𝑥
− 1)4
=𝑒4𝑥
− 4𝑒3𝑥
+ 6𝑒2𝑥
− 4𝑒 𝑥
+ 1
∴The sequence ans= Co.eff of
𝑥11
11!
in E(x)
=411
− 4(311
)+6(211
)-4(111
)=3,498,000
7) Determine the sequences generated by each of the following
exponential generating functions.
i)E(x) = 4𝑒4𝑥
ii)E(x)=2𝑒 𝑥
+ 3(𝑥2
)
iii)E(x)=8𝑒7𝑥
− 3𝑒4𝑥
iv)E(x)=𝑒3𝑥
− 28𝑥3
− 6𝑥2
+ 9

6. Solution:
i)E(x)=4∑
(4𝑥) 𝑛
𝑛!
∞
𝑛=0 ⇒ {4,42
,43
,44
,…}
ii)E(x)=8∑
(7𝑥) 𝑛
𝑛!
∞
𝑛=0 -3∑
(4𝑥) 𝑛
𝑛!
∞
𝑛=0
Seq =87 𝑛
− 34 𝑛
, n=0,1,2,…
iii)E(x)=2∑
𝑥 𝑛
𝑛!
∞
𝑛=0 + 3𝑥2
E(x)=2(1+
𝑥
1!
+
𝑥2
2!
+
𝑥3
3!
+…)+3𝑥2
Seq={2,2,(2+3),2,…}
Seq={2,2,5,2,…}
iv) E(x)=𝑒3𝑥
− 28𝑥3
− 6𝑥2
+ 9𝑥
E(x)=∑
3 𝑛 𝑥 𝑛
𝑛!
− 28𝑥3
− 6𝑥2
+ 9∞
𝑛=0 𝑥
⇒Seq={30
,31
+ 9,32
− 6,33
− 28,34
,3 5
,36
,…}
Seq={1,12,3,-1, 34
,3 5
,36
,…}