Math Assignment: Combinations, Derivatives, and Binomial Expansion

This assignment presents solutions to a set of mathematical problems. The first problem involves calculating the number of possible four-letter codes formed from a set of eight letters, allowing for repetition. The second problem requires calculating the number of ways to select five dresses from a set of fifty-two dresses, framed as a combination problem. The third problem focuses on finding the stationary point and minimum value of a quadratic function using differentiation. The solution includes finding the first and second derivatives to determine the nature of the stationary point. Finally, the assignment addresses the binomial expansion of (x + 3)^5, calculating the coefficient of x^4 using the binomial theorem.

14)
Answer:
Four letters are to be selected out of:
A, B, C, D, E, F, G, H
1

that is total 8 letters.
To form a code of four letters,the first letter can be selected in 8 different
ways.
Since repetition is allowed, the second letter of the code can also be selected
in 8 different ways from the set of available letters.
Similarly, the third and fourth letter for the code can each be selected in 8
different ways.
Therefore, the total number of ways a four letter code cab be formed from
the set of eight letters is:
8 × 8 × 8 × 8 = 4096
15)
Answer:
Since the order of cards in not important, it is combination counting prob-
lem.
5 dresses can be selected from a total 52 dresses in52
5 ways:
52
5 = 52!
47! × 5!
= 2, 598, 960
16)
f (x) = 2x2 + 12x
Answer:
Stationary point of f (x) is given by:
df
dx = 0
Now, the derivative of f (x) is:
d
dx(2x2 + 12x) = 4x + 12
Stationary point therefore is:
4x + 12 = 0 =⇒ x = −3
To determine whether this point is a maxima or minima, the sign of second
derivative at the point has to be determined.If:
d2f
dx2 x=a > 0 =⇒ a is minima
2

d2f
dx2 x=a < 0 =⇒ a is maxima
Now,
d2f
dx2 x=−3 = d
dx(4x + 12)
x=−3 = 4
Therefore, x = −3 is a minima of f (x).The minimum value of f (x) is:
f(−3) = 2x2 + 12x−3 = 2(−3)2 + 12(−3) = −18
17
Coefficient of x4
Answer:
The Binomial expansion of (ax + b)n is:
(ax + b)n =
nX
k=0
n
k an−kxn−kbk
where, n
k an−kbk is the coefficient ofxn−k. The binomialexpansion of
(x + 3)5 is:
(x + 3)5 =
5X
k=0
5
k x5−k3k
for k = 1, the power of x is 4.Thus, when k = 1, the coefficient of x4 is:
C1 = 5
1 31
5
1 = 5!
1! × 4!
= 5
31 = 3
Therefore,
C4 = 5 × 3 = 15
3