WAEC SSCE - Further Mathematics - 2012
Ajụjụ 1 Ripọtì
Which of the following is nor a measure of central tendency?
Akọwa Nkọwa
Variance is not a measure of central tendency. Measures of central tendency are used to describe the typical or central value of a set of data, while variance is a measure of how spread out the data is from the mean. Variance is a measure of variability, not centrality. Therefore, the answer is "Variance."
Ajụjụ 2 Ripọtì
If the midpoint of the line joining (1 - k, -4) and (2, k + 1) is (-k, k), find the value of k.
Akọwa Nkọwa
We can start by using the midpoint formula which states that the midpoint of a line joining two points \((x_1, y_1)\) and \((x_2, y_2)\) is \((\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})\). So, the midpoint of the line joining (1-k, -4) and (2, k+1) is: \begin{align*} &\left(\frac{(1-k)+2}{2},\frac{(-4)+(k+1)}{2}\right) \\ &\Rightarrow \left(\frac{3-k}{2},\frac{k-3}{2}\right) \end{align*} We are given that the midpoint is (-k, k), so we can equate the x and y coordinates: \begin{align*} \frac{3-k}{2} &=-k \\ \frac{k-3}{2} &=k \end{align*} Solving for k, we get: \begin{align*} k &= 2 \\ \end{align*} Therefore, the value of k is 2, which is the answer option labeled as "-2".
Ajụjụ 3 Ripọtì
Which of the following sets is equivalent to \((P \cup Q) \cap (P \cup Q')\)?
Akọwa Nkọwa
Ajụjụ 6 Ripọtì
If (x + 1) is a factor of the polynomial \(x^{3} + px^{2} + x + 6\). Find the value of p.
Akọwa Nkọwa
If (x + 1) is a factor of the polynomial, it means that if we substitute -1 in the polynomial, it should give us zero. Therefore: \((-1)^3 + p(-1)^2 -1 + 6 = 0\) Simplifying the above equation, we get: \(-1 + p + 5 = 0\) \(p = -4\) Hence, the value of p is -4. So the correct answer is.
Ajụjụ 7 Ripọtì
Evaluate \(\log_{10}(\frac{1}{3} + \frac{1}{4}) + 2\log_{10} 2 + \log_{10} (\frac{3}{7})\)
Akọwa Nkọwa
To simplify this expression, we can first use the identity: $$\log_{a}(b) + \log_{a}(c) = \log_{a}(bc)$$ Using this identity, we can simplify the given expression as follows: \begin{align*} \log_{10}\left(\frac{1}{3}+\frac{1}{4}\right) + 2\log_{10}(2) + \log_{10}\left(\frac{3}{7}\right) &= \log_{10}\left(\frac{7}{12}\right) + \log_{10}(2^2) + \log_{10}\left(\frac{3}{7}\right) \\ &= \log_{10}\left(\frac{7}{12} \cdot 2^2 \cdot \frac{3}{7}\right) \\ &= \log_{10}(1) \\ &= 0 \end{align*} Therefore, the answer is 0.
Ajụjụ 8 Ripọtì
Find the values of x at the point of intersection of the curve \(y = x^{2} + 2x - 3\) and the lines \(y + x = 1\).
Akọwa Nkọwa
To find the point of intersection of the curve and the line, we need to solve the system of equations formed by equating the two equations: \begin{align*} y &= x^2 + 2x - 3 \\ y &= -x + 1 \end{align*} Setting the right-hand sides equal to each other, we get: \begin{align*} x^2 + 2x - 3 &= -x + 1 \\ x^2 + 3x - 4 &= 0 \\ (x + 4)(x - 1) &= 0 \end{align*} Thus, the values of x at the points of intersection are x = -4 and x = 1. To find the corresponding y-values, we substitute these values of x back into either equation. Using the equation y = x^2 + 2x - 3, we get: \begin{align*} y &= (-4)^2 + 2(-4) - 3 \\ &= 7 \end{align*} and \begin{align*} y &= (1)^2 + 2(1) - 3 \\ &= 0 \end{align*} Therefore, the points of intersection are (-4, 7) and (1, 0). So, the correct answer is (1, -4).
Ajụjụ 9 Ripọtì
A stone is dropped from a height of 45m. Find the time it takes to hit the ground. \([g = 10 ms^{-2}]\)
Akọwa Nkọwa
To solve this problem, we can use the formula: \[ s = ut + \frac{1}{2}at^2 \] where s is the distance, u is the initial velocity, t is the time, and a is the acceleration due to gravity. In this case, the initial velocity is zero because the stone is dropped from rest. We also know that the distance is 45m and the acceleration due to gravity is 10 \(ms^{-2}\). Thus, we have: \begin{align*} s &= ut + \frac{1}{2}at^2 \\ 45 &= 0t + \frac{1}{2}(10)t^2 \\ 45 &= 5t^2 \\ t^2 &= 9 \\ t &= 3 \end{align*} Therefore, the time it takes for the stone to hit the ground is 3 seconds. Hence, the answer is (a) 3.0 seconds.
Ajụjụ 10 Ripọtì
If \(y = x^{3} - x^{2} - x + 6\), find the values of x at the turning point.
Akọwa Nkọwa
Ajụjụ 12 Ripọtì
The distance s in metres covered by a particle in t seconds is \(s = \frac{3}{2}t^{2} - 3t\). Find its acceleration.
Akọwa Nkọwa
To find the acceleration, we need to differentiate the distance formula with respect to time (t): \begin{align*} s &= \frac{3}{2}t^{2} - 3t \\ \frac{d}{dt}s &= \frac{d}{dt}\left(\frac{3}{2}t^{2}\right) - \frac{d}{dt}(3t) \\ \frac{d}{dt}s &= 3t - 3 \\ \end{align*} Therefore, the acceleration is the second derivative of the distance formula with respect to time: \begin{align*} \frac{d^{2}}{dt^{2}}s &= \frac{d}{dt}(3t - 3) \\ \frac{d^{2}}{dt^{2}}s &= 3 \\ \end{align*} Thus, the acceleration of the particle is a constant value of 3 \(ms^{-2}\). Therefore, the answer is \(3 ms^{-2}\).
Ajụjụ 13 Ripọtì
A box contains 4 red and 3 blue identical balls. If two are picked at random, one after the other without replacement, find the probability that one is red and the other is blue.
Akọwa Nkọwa
Ajụjụ 14 Ripọtì
If P(x - 3) + Q(x + 1) = 2x + 3, find the value of (P + Q).
Akọwa Nkọwa
To find the value of (P + Q), we need to first expand the left-hand side of the equation using distributive property. P(x - 3) + Q(x + 1) = Px - 3P + Qx + Q Then we can simplify it by combining the like terms. Px + Qx - 3P + Q + 2x + 3 = 0 Now, we can group the like terms together: (P + Q)x - 3P + Q + 3 = 2x + 3 Since the coefficients of x on both sides of the equation are equal, we can equate the corresponding coefficients of x: (P + Q) = 2 Therefore, the value of (P + Q) is 2. So the correct answer is: - 2
Ajụjụ 15 Ripọtì
A fair die is tossed twice. Find the probability of obtaining a 3 and a 5.
Akọwa Nkọwa
When a die is tossed, there are six possible outcomes, each with equal probability. Therefore, the probability of getting a 3 on the first toss is 1/6, and the probability of getting a 5 on the second toss is also 1/6. Since we want both events to occur, we need to multiply their probabilities: \[\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}\] So, the probability of obtaining a 3 and a 5 is 1/36. Therefore, the correct option is: - \(\frac{1}{36}\)
Ajụjụ 16 Ripọtì
Simplify \(\frac{x^{3n + 1}}{x^{2n + \frac{5}{2}}(x^{2n - 3})^{\frac{1}{2}}}\)
Akọwa Nkọwa
Ajụjụ 17 Ripọtì
Differentiate \(\frac{x}{x + 1}\) with respect to x.
Akọwa Nkọwa
To differentiate \(\frac{x}{x + 1}\) with respect to x, we can use the quotient rule of differentiation, which states that for functions u(x) and v(x), the derivative of \(\frac{u(x)}{v(x)}\) is given by: \[\frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x) v(x) - u(x) v'(x)}{(v(x))^2}\] Applying this rule to the given function, we have: \[u(x) = x\] \[v(x) = x + 1\] So, we need to find u'(x) and v'(x): \[u'(x) = 1\] \[v'(x) = 1\] Substituting these values into the quotient rule formula, we get: \[\frac{d}{dx} \left( \frac{x}{x + 1} \right) = \frac{1(x + 1) - x(1)}{(x + 1)^2}\] Simplifying the numerator and denominator, we get: \[\frac{d}{dx} \left( \frac{x}{x + 1} \right) = \frac{1}{(x + 1)^2}\] Therefore, the correct answer is \(\frac{1}{(x + 1)^2}\).
Ajụjụ 18 Ripọtì
The equation of a circle is \(3x^{2} + 3y^{2} + 24x - 12y = 15\). Find its radius.
Akọwa Nkọwa
To find the radius of the circle, we need to use the standard form of the equation of a circle, which is \((x - a)^{2} + (y - b)^{2} = r^{2}\), where \((a, b)\) is the center of the circle and \(r\) is the radius. To convert the given equation to standard form, we can complete the square for both \(x\) and \(y\): \begin{align*} 3x^{2} + 3y^{2} + 24x - 12y &= 15 \\ 3(x^{2} + 8x) + 3(y^{2} - 4y) &= 15 \\ 3(x^{2} + 8x + 16) + 3(y^{2} - 4y + 4) &= 15 + 3(16) + 3(4) \\ 3(x + 4)^{2} + 3(y - 2)^{2} &= 72 \\ (x + 4)^{2} + (y - 2)^{2} &= 8^{2} \end{align*} Comparing this with the standard form, we see that the center of the circle is \((-4, 2)\) and the radius is \(8\). Therefore, the answer is (d) 5.
Ajụjụ 19 Ripọtì
A polynomial is defined by \(f(x + 1) = x^{3} + px^{2} - 4x + 2\), find f(2).
Akọwa Nkọwa
Ajụjụ 20 Ripọtì
Given that \(\sqrt{6}, 3\sqrt{2}, 3\sqrt{6}, 9\sqrt{2},...\) are the first four terms of an exponential sequence (G.P), find in its simplest form the 8th term.
Akọwa Nkọwa
Ajụjụ 22 Ripọtì
QRS is a triangle such that \(\overrightarrow{QR} = (3i + 2j)\) and \(\overrightarrow{SR} = (-5i + 3j)\), find \(\overrightarrow{SQ}\).
Akọwa Nkọwa
To find \(\overrightarrow{SQ}\), we can use the fact that \(\overrightarrow{SR} = \overrightarrow{SQ} + \overrightarrow{QR}\). Rearranging this equation to solve for \(\overrightarrow{SQ}\) gives us: $$\overrightarrow{SQ} = \overrightarrow{SR} - \overrightarrow{QR} = (-5i + 3j) - (3i + 2j) = -8i + j$$ Therefore, the value of \(\overrightarrow{SQ}\) is -8i + j. So, the correct option is (A) 8i + j.
Ajụjụ 23 Ripọtì
Two forces 10N and 6N act in the directions 060° and 330° respectively. Find the x- component of their resultant.
Akọwa Nkọwa
Ajụjụ 24 Ripọtì
If r denotes the correlation coefficient between two variables, which of the following is always true?
Akọwa Nkọwa
The correct answer is: \(-1 \leq r \leq 1\). The correlation coefficient (r) measures the degree of linear association between two variables. The value of r ranges from -1 to 1, where -1 indicates a perfect negative correlation, 0 indicates no correlation, and 1 indicates a perfect positive correlation. Since r can take on any value between -1 and 1 (including -1 and 1), the correct statement is \(-1 \leq r \leq 1\), which means that the correlation coefficient is always between -1 and 1, inclusive. Therefore, options (A), (B), and (C) are incorrect.
Ajụjụ 25 Ripọtì
A straight line makes intercepts of -3 and 2 on the x- and y- axes respectively. Find the equation of the line.
Akọwa Nkọwa
Ajụjụ 26 Ripọtì
In computing the mean of 8 numbers, a boy mistakenly used 17 instead of 25 as one of the numbers and obtained 20 as the mean. Find the correct mean
Akọwa Nkọwa
Let's call the sum of the 8 correct numbers "S" and the incorrect number that was added "x". The mean of the 8 numbers is: S/8 But the boy used 17 instead of 25, so the sum he used was: S + 17 - 25 = S - 8 And he got a mean of 20, so we can set up the equation: (S - 8)/8 = 20 Solving for S, we get: S = 168 So the correct mean is: S/8 = 168/8 = 21 Therefore, the correct mean is 21, which is.
Ajụjụ 27 Ripọtì
Find the constant term in the binomial expansion of \((2x - \frac{3}{x})^{8}\).
Akọwa Nkọwa
Ajụjụ 28 Ripọtì
Given that \(P = \begin{pmatrix} 2 & 1 \\ 5 & -3 \end{pmatrix}\) and \(Q = \begin{pmatrix} 4 & -8 \\ 1 & -2 \end{pmatrix}\), Find (2P - Q).
Akọwa Nkọwa
To find the value of (2P - Q), we first need to compute 2P and Q, and then subtract Q from 2P. To compute 2P, we multiply each element of matrix P by 2: 2P = \(\begin{pmatrix} 4 & 2 \\ 10 & -6 \end{pmatrix}\) To subtract Q from 2P, we subtract each corresponding element of matrix Q from matrix 2P: 2P - Q = \(\begin{pmatrix} 4-4 & 2+8 \\ 10-1 & -6+2 \end{pmatrix}\) = \(\begin{pmatrix} 0 & 10 \\ 9 & -4 \end{pmatrix}\) Therefore, the answer is \(\begin{pmatrix} 0 & 10 \\ 9 & -4 \end{pmatrix}\).
Ajụjụ 29 Ripọtì
From the diagram above, which of the following represents the vector V in component form?
Akọwa Nkọwa
Ajụjụ 30 Ripọtì
A binary operation, \(\Delta\), is defined on the set of real numbers by \(a \Delta b = a + b + 4\). Find the identity element.
Akọwa Nkọwa
An identity element in a binary operation is an element such that when it operates with any other element of the set, it does not change the other element. In this case, we need to find an element, say "x," such that for any real number "a," $$a \Delta x = a$$ Substituting the given definition of the operation, we get: $$a + x + 4 = a$$ Solving for x, we get: $$x = -4$$ Thus, -4 is the identity element for the given binary operation.
Ajụjụ 32 Ripọtì
Find the unit vector in the direction of the vector \(-12i + 5j\).
Akọwa Nkọwa
To find the unit vector in the direction of the vector \(-12i + 5j\), we need to divide the vector by its magnitude. The magnitude of a vector with components \(a\) and \(b\) is given by the formula \(\sqrt{a^2+b^2}\). So, the magnitude of the vector \(-12i + 5j\) is \(\sqrt{(-12)^2+5^2} = 13\). Now, to get the unit vector, we divide each component of the vector by its magnitude: \[\frac{-12}{13}i + \frac{5}{13}j\] This is the unit vector in the direction of the vector \(-12i + 5j\). Therefore, the correct option is \(\frac{-12i}{13} + \frac{5j}{13}\).
Ajụjụ 33 Ripọtì
Find the acute angle between the lines 2x + y = 4 and -3x + y + 7 = 0.
Akọwa Nkọwa
To find the acute angle between two lines, we need to find the angle between their respective direction vectors. The direction vector of the line 2x + y = 4 is v = i + 2j and the direction vector of the line -3x + y + 7 = 0 is w = -3i + j. The acute angle θ between two vectors a and b is given by the formula cos(θ) = (a.b) / (|a|.|b|), where a.b is the dot product of the vectors and |a| and |b| are their respective magnitudes. Using this formula, we can find cos(θ) for the given direction vectors as follows: cos(θ) = (v.w) / (|v|.|w|) = (-5) / (√5.√10) = -1/√2 Since we are looking for the acute angle, we need to take the inverse cosine of -1/√2 in the range [0, π/2]: θ = cos-1(-1/√2) ≈ 45° Therefore, the acute angle between the given lines is approximately 45°, which is.
Ajụjụ 34 Ripọtì
The diagram above is a velocity- time graph of a moving object. Calculate the distance travelled when the acceleration is zero.
Akọwa Nkọwa
Ajụjụ 36 Ripọtì
Given that \(^{n}P_{r} = 90\) and \(^{n}C_{r} = 15\), find the value of r.
Akọwa Nkọwa
We know that: $$^{n}P_{r} = \frac{n!}{(n-r)!} = 90$$ and $$^{n}C_{r} = \binom{n}{r} = \frac{n!}{r!(n-r)!} = 15$$ To find the value of r, we can use the formula: $$^{n}C_{r} = \frac{^{n}P_{r}}{r!}$$ Substituting the given values, we get: $$15 = \frac{90}{r!}$$ Simplifying the equation, we get: $$r! = 6$$ The only integer value of r that satisfies this equation is 3, since 3! = 6. Therefore, the answer is r = 3.
Ajụjụ 37 Ripọtì
The marks obtained by 10 students in a test are as follows: 3, 7, 6, 2, 8, 5, 9, 1, 4 and 10. Find the variance.
Akọwa Nkọwa
Ajụjụ 38 Ripọtì
Find the number of different arrangements of the word IKOTITINA.
Ajụjụ 39 Ripọtì
The angle of a sector of a circle is 0.9 radians. If the radius of the circle is 4cm, find the length of the arc of the sector.
Akọwa Nkọwa
The formula for finding the length of an arc of a sector is given by L = rθ, where L is the length of the arc, r is the radius of the circle, and θ is the angle in radians. Using this formula and the given values, we have: L = 4 x 0.9 L = 3.6 cm Therefore, the length of the arc of the sector is 3.6 cm. Answer: 3.6 cm.
Ajụjụ 40 Ripọtì
The marks obtained by 10 students in a test are as follows: 3, 7, 6, 2, 8, 5, 9, 1, 4 and 10. Find the mean mark.
Akọwa Nkọwa
To find the mean mark, we add up all the marks and divide by the number of students. Adding up all the marks, we get: 3 + 7 + 6 + 2 + 8 + 5 + 9 + 1 + 4 + 10 = 55 There are 10 students, so we divide the sum of the marks by 10: 55 / 10 = 5.50 Therefore, the mean mark is 5.50. Hence, the correct option is: 5.50.
Ajụjụ 41 Ripọtì
(a) Evaluate : \(\int_{1} ^{4} \frac{x(3x - 2)}{2\sqrt{x}} \mathrm {d} x\)
(b) The equation of a circle is given by \(2x^{2} + 2y^{2} - 8x + 5y - 10 = 0\). Find the :
(i) coordinates of the centre ; (ii) radius of the circle .
Ajụjụ 42 Ripọtì
(a) Find the angle between the vectors \(a = \begin{pmatrix} -3 \\ 4 \end{pmatrix}\) and \(b = \begin{pmatrix} -8 \\ -15 \end{pmatrix}\).
(b) Given that \(a = (4N, 060°)\) and \(b = (3N, 120°)\), find, in component form, the unit vector along \(a - b\).
Ajụjụ 43 Ripọtì
(a) Write down the matrix A of the linear transformation \(A(x, y) \to (2x -y, -5x + 3y)\).
(b) If \(B = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}\), find :
(i) \(A^{2} - B^{2}\) ; (ii) matrix \(C = B^{2} A\) ; (iii) the point \(M(x, y)\) whose image under the linear transformation \(C\) is \(M' (10, 18)\).
(c) What is the relationship between matrix A and matrix C?
Ajụjụ 44 Ripọtì
Express \(3x^{2} - 6x + 10\) in the form \(a(x - b)^{2} + c\), where a, b and c are integers. Hence state the minimum value of \(3x^{2} - 6x + 10\) and the value of x for which it occurs.
Ajụjụ 45 Ripọtì
(a)(i) Find the sum of the series \(A(1 + r) + A(1 + r)^{2} + ... + A(1 + r)^{n}\).
(ii) Given that r = 8% and A = GH 40.00, find the sum of the 6th to 10th terms of the series in (i).
(b) Find the equation of the tangent to the curve \(y = \frac{1}{x}\) at the point on the curve when x = 2.
Ajụjụ 46 Ripọtì
Two functions g and h are defined on the set R of real numbers by \(g : x \to x^{2} - 2\) and \(h : x \to \frac{1}{x + 2}\). Find :
(a) \(h^{-1}\), the inverse of h ;
(b) \(g \circ h\), when \(x = -\frac{1}{2}\).
Ajụjụ 47 Ripọtì
(a) A fair die with six faces is thrown six times. Calculate, correct to three decimal places, the probability of obtaining :
(i) exactly three sixes ; (ii) at most three sixes.
(b) Eight percent of screws produced by a machine are defective. From a random sample of 10 screws produced by the machine, find the probability that :
(i) exactly two will be defective ; (ii) not more than two will be defective.
Ajụjụ 49 Ripọtì
(a) The polynomial \(f(x) = x^{3} + px^{2} - 10x + q\) is exactly divisible by \(x^{2} + x - 6\). Find the :
(i) values of p and q ; (ii) third factor.
(b) The volume of a cube is increasing at the rate of \(2\frac{1}{2} cm^{3} s^{-1}\). Find the rate of change of the side of the base when its length is 2cm.
Ajụjụ 50 Ripọtì
Three forces \(-63j , 32.14i + 38.3j\) and \(14i - 24.25j\) act on a body of mass 5kg. Find, correct to one decimal place, the :
(a) magnitude of the resultant force ;
(b) acceleration of the body.
Ajụjụ 51 Ripọtì
The table gives the distribution of heights in metres of 100 students.
| Height | 1.40-1.42 | 1.43-1.45 | 1.46-1.48 | 1.49-1.51 | 1.52-1.54 | 1.55-1.57 | 1.58-1.60 | 1.61-1.63 |
| Freq | 2 | 4 | 19 | 30 | 24 | 14 | 6 | 1 |
(a) Calculate the : (i) mean height ; (ii) mean deviation of the distribution.
(b) What is the probability that the height of a student selected at random is greater than the mean height of the distribution?
Ajụjụ 52 Ripọtì
(a) Given that \(p = (4i - 3j)\) and \(q = (-i + 5j)\), find r such that \(|r| = 15\) and is in the direction \((2p + 3q)\).
(b) 
Forces of magnitude 8N, 6N and 4N act at the point P, as shown in the above diagram. Find the : (i) magnitude ; (ii) direction of the resultant force.
Ajụjụ 53 Ripọtì
Write down the first three terms of the binomial expansion \((1 + ax)^{n}\) in ascending powers of x. If the coefficients of x and x\(^{2}\) are 2 and \(\frac{3}{2}\) respectively, find the values of a and n.
Ajụjụ 54 Ripọtì
The marks scored by 35 students in a test are given in the table below.
| Marks | 1-5 | 6-10 | 11-15 | 16-20 | 21-25 | 26-30 |
| Frequency | 2 | 7 | 12 | 8 | 5 | 1 |
Draw a histogram for the distribution.
Ajụjụ 55 Ripọtì
(a) Two items are selected at random from four items labelled (p, q, r, s).
(i) List the sample space if sampling is done (1) with replacement ; (2) without replacement.
(ii) Find the probability that r is at least one of the two objects selected : (1) in a(i)1 ; (2) in a(i)2.
(b) How many whole numbers from 100 to 999 are divisible by (i) 4 ; (ii) both 3 and 4?
Akọwa Nkọwa
None
Ajụjụ 56 Ripọtì
The twenty-first term of an Arithmetic Progression is \(5\frac{1}{2}\) and the sum of the first twenty-one terms is \(94\frac{1}{2}\). Find the :
(a) first term ; (b) common difference ; (c) sum of the first thirty terms.
Ajụjụ 57 Ripọtì
(a) A body P of mass 5kg is suspended by two light inextensible strings AP and BP attached to a ceiling. If the strings are inclined at angles 40° and 30° respectively to the downward vertical, find the tension in each of the strings. [Take \(g = 10 ms^{-2}\)].
(b) A constant force F acts on a toy car of mass 5 kg and increases its velocity from 5 ms\(^{-1}\) to 9 ms\(^{-1}\) in 2 seconds. Calculate :
(i) the magnitude of the force ; (ii) velocity of the toy car 3 seconds after attaining a velocity of 9 ms\(^{-1}\).
Ajụjụ 58 Ripọtì
Simplify \(^{n + 1}C_{4} - ^{n - 1}C_{4}\)
= \(\frac{(n + 1)!}{4! (n - 3)!} - \frac{(n - 1)!}{4! (n - 5)!}\)
= \(\frac{(n + 1)(n)(n - 1)(n - 2)(n - 3)!}{4! (n - 3)!} - \frac{(n - 1)(n - 2)(n - 3)(n - 4)(n - 5)!}{4! (n - 5)!}\)
= \(\frac{(n + 1)(n)(n - 1)(n - 2)}{4!} - \frac{(n - 1)(n - 2)(n - 3)(n - 4)}{4!}\)
= \(\frac{(n - 1)(n - 2) [n(n + 1) - (n - 3)(n - 4)]}{4!}\)
= \(\frac{(n - 1)(n - 2) [n^{2} + n - n^{2} + 7n - 12]}{24}\)
= \(\frac{(n - 1)(n - 2)[8n - 12]}{24}\)
= \(\frac{(n - 1)(n - 2)(2n - 3)}{6}\)
