Math Assignment: Combinations, Equations, and Financial Calculations

This mathematics assignment presents solutions to a variety of problems. Task A explores combinations, differentiating them from permutations and applying the correct formula to calculate the number of possible groups. Task B analyzes data related to tile dimensions and prices, explaining the meaning of the provided summations. Task C delves into demand and supply equations, determining the dependent and independent variables, calculating the slope, and finding the equilibrium point. Task D involves linear programming, graphically representing inequalities, identifying the feasible region, and maximizing an objective function. Finally, Task E tackles financial calculations, including compound interest, loan repayments, and percentage-based price changes. The assignment demonstrates the application of various mathematical concepts to real-world scenarios.

Using and managing data and
information

TABLE OF CONTENTS
TASK A...........................................................................................................................................3
TASK B...........................................................................................................................................3
TASK C...........................................................................................................................................4
TASK D...........................................................................................................................................5
TASK E...........................................................................................................................................7
REFERENCES................................................................................................................................9

TASK A
1. She needs to find the combinations in this case because there is no definite order in which
groups must be made. If any order was given, then permutations would be used but the
groups are to be made of 5 and the arrangement is not necessary but only selecting 5
students from 20 and making possible groups of 5 is needed. The combinations are used
when the selection of objects or the things has to be done when the selection order is
immaterial while the permutations are used when the objects needs to be arranged in
definite order (Lockwood, Wasserman and McGuffey, 2018).
2. The above formula which is 20! / (20-5)! is wrong because it will give the permutations
which is nPr. This will give the number of permutations of 20 students taken 5 at a time.
We need to find the combinations in which 20 students can be grouped into groups of 5.
We are not interested about the arrangement in which they will be ordered. Therefore,
this formula is wrong (Muntean and et.al., 2018).
3. The correct formula to find the combinations or the selection of r objects from the n
objects which will be denoted by nCr is as given below:
nCr = n!/(r!)(n-r)!
In order to find the combinations in this case, the above formula can be used as follows:
20C5 = 20! /5! (20-5)!
= (20*19*18*17*16*15*14*....*1) / (5!)(15!)
= {(20*19*18*17*16)* (15*14*....*1)} / {(5*4*3*2*1) (15!)}
= {(20*19*18*17*16)* (15!)} / {(5*4*3*2*1) * (15!)}
= (19*18*17*16) / (3*2*1)
= 19*3*17*16
= 15504
TASK B
Sizes Dimension in
cm (x)
Unit price in
pound (y)
x2 xy
1 3 0.7 9 2.1
2 5 0.9 25 4.5
3 6 1 36 6
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4 8 1.2 64 9.6
5 9 1.5 81 13.5
6 10 1.8 100 18
7 12 2 144 24
1. ∑x2 = 9+25+36+64+81+100+144 = 459
2. Since the tile is in shape of the square and the side of the square is denoted by x
which means x2 will be the area of one tile as the area of square is (side)2. But this is
∑x2 which will means the total area of all the tiles having sizes 1, 2, ....., 7. Therefore,
∑x2 is the total area of all the tiles (Deltell, 2020).
3. ∑x y = 2.1+4.5+6+9.6+13.5+ 18+24 = 77.7
4. Since x is dimension in centimetre and y is the price of the tile in pound then, ∑x y
will be the sum of the product of dimension and the price of 1 tile of the respective
size.
TASK C
1. The equation, p = aQ+b, is of the form y = mx+b where y is the dependent variable, m is
the slope and b is the y intercept.
So, according to the equation given above p is the dependent variable.
In order to make p as the independent variable the equation turns out to be
Q = a + bP
Here, Q is the dependent variable which depends on the independent variable P. a is the
Q intercept or the x-intercept. This is the quantity demanded when the price = 0 which is
also known as autonomous demand. It is independent of Price. b is the inverse of slope as
the change in the quantity is to be measured with particular change in price. So b will be
the slope which is change in quantity over change in price (Lubis and Nasution, 2017).
b = change in quantity / change in price
From the graph given, P1= 20 and Q1 = 20 whereas P2 = 40 and Q2 = 10
So, b = (Q2 - Q1)/ (P2 - P1)
= (10–20) / (40-20)
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= -1/2
= - 0.5
Therefore, the equation becomes Q = a + (-0.5P)
The value of b is always negative because there is an inverse relationship between the
price and the quantity demanded.
The value of a is 30 because the quantity demanded at price = 0 is 30.
Therefore, the equation will be Q = 30-0.5P
If the market price is £40, then the quantity demanded will be:
Q = 30-0.5P
Q = 30 – 0.5 (40)
Q = 30 – 20
Q = 10
Therefore, the quantity demanded will be 10.
2. The name of the point A in the graph having supply and demand functions is Equilibrium
point. This is the point at which the quantity demanded equals to the quantity supplied.
Therefore, the equilibrium price is 20 and the equilibrium quantity is also 20.
TASK D
The graph for the following equations is given below:
3A+6B<=300
4A+2B<=160
A>=0
B>=0
The objective function is P = 2A+2B
5

The feasible region is made by joining the points (0,50), (20,40) and (40,0) (Abdel-Basset and
et.al., 2019)
The value of the objective function for point (0, 50) is
P = 2A+2B
P = 2*0+2*50
= 100
The value of the objective function for point (40,0) is
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P = 2A+2B
P = 2*40+2*0
= 80
The value of the objective function for point (20,40) is
P = 2A+2B
P = 2*20+2*40
= 120
The values of A and B for which the value of objective function is maximum is (20, 40) and the
maximum value is 120.
TASK E
1. In order to solve this, the formula for compound interest can be used. The compound
interest formula basically helps in estimating how much can be earned with the savings
account. Calculating this is little bit complex as it does not only include the annual
interest rate and the number of years but also number of times the interest is compounded
in a year.
The formula which can be used is as follows:
FV = P (1+r/m) ^ mt (Melchor and Sáez, 2019)
Here, FV = Future Value of the investment
P = Initial balance
r = Annual interest rate
m = the number of times in which the interest is compounded per year.
t = the number of years for which the money is invested for.
Putting the values given in the above mentioned formula, we get
FV = ?
P = £20,000
r = 2%
m = 12 months
t = 5 years
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FV = P (1+r/m) ^ mt (AbuEloun and Abu-Naser, 2017)
= 20,000 (1+2/12) ^ 12 * 5
= 20,000 (1+1/6) ^ 60
= 20,000 (7/6) ^ 60
= (140,000/6) ^ 60
= 23,333.33 ^ 60
= £22,101.50
Compound interest = £2,101.58.
2. The formula which can be used to find the repayments is as given below:
Payment = P x (r / n) x (1 + r / n)^n(t)] / (1 + r / n)^n(t) – 1 (Eastaway, 2021)
Here, P = loan amount or principal
r = Annual interest rate on the loan
t = Number of years in which the loan has to be repaid which is also known as term.
n = The number of payments per year as in case of monthly payments it is 12.
= £200,000 x (3.6 / 1) x (1 + 3.6/ 1)^1(25) / (1 + 3.6 / 25)^ 1(25) – 1
= 200,000 x (3.6) x (4.6)^(25)/ (1 + 3.6 / 25)^ (25) – 1
= 720,000 x 3.7 x 1016 / 1.144 ^ 25 – 1
= 2664000 x 1016 / 28.88-1
= 2664000 x 1016 / 27.88
= £95,552.36 x 1016
3. Cost of Clementina orange ceramic vase down by 25% = £16.50
Let the original cost of the vase is x, then the equation becomes
x – 25% of x = £16.50
x – (25/100 * x) = £16.50
100x – 25x = 1650
75x = 1650
x = 1650/75
x= 22.
Therefore, the price of the vase before the sale was £22.
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REFERENCES
Books and Journals
Abdel-Basset, M. and et.al., 2019. A novel method for solving the fully neutrosophic linear
programming problems. Neural Computing and Applications. 31(5). pp.1595-1605.
AbuEloun, N.N. and Abu-Naser, S.S., 2017. Mathematics intelligent tutoring system.
Deltell, A.F., 2020. From statistics to data science. A review of the data science degrees in
Spanish public universities. BEIO, Boletín de Estadística e Investigación
Operativa. 36(1). pp.77-91.
Eastaway, R., 2021. Maths on the back of an envelope: Clever ways to (roughly) calculate
anything.
Lockwood, E., Wasserman, N.H. and McGuffey, W., 2018. Classifying combinations:
Investigating undergraduate students’ responses to different categories of combination
problems. International Journal of Research in Undergraduate Mathematics
Education. 4(2). pp.305-322.
Lubis, A. and Nasution, A.A., 2017. How Do Higher-Education Students Use Their Initial
Understanding to Deal with Contextual Logic-Based Problems in Discrete
Mathematics?. International Education Studies. 10(5). pp.72-86.
Melchor, C.M. and Sáez, M.P., 2019. Apps for teaching maths of finance: An empirical
approach. Anales de ASEPUMA. (27). p.7.
Muntean, C.H. and et.al., 2018, October. Investigating the impact of an immersive computer-
based math game on the learning process of undergraduate students. In 2018 IEEE
Frontiers in Education Conference (FIE) (pp. 1-8). IEEE.
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