WASSCE Further Maths: Combinatorics, Sequences & Matrices Practice Questions & Answers

Let the first term be $a$a and the common difference be $d$d. The formula for the nth term is $U_n = a + (n-1)d$Un​=a+(n−1)d. We have $U_4 = a + 3d = 10$U4​=a+3d=10 and $U_7 = a + 6d = 19$U7​=a+6d=19. Subtracting the first equation from the second gives: $(a + 6d) - (a + 3d) = 19 - 10 \Rightarrow 3d = 9 \Rightarrow d = 3$(a+6d)−(a+3d)=19−10⇒3d=9⇒d=3.

Expanding along the first row: Det $= 1(1(0) - 4(6)) - 2(0(0) - 4(5)) + 3(0(6) - 1(5))$=1(1(0)−4(6))−2(0(0)−4(5))+3(0(6)−1(5)). Thus, Det $= 1(0 - 24) - 2(0 - 20) + 3(0 - 5) = -24 + 40 - 15 = 1$=1(0−24)−2(0−20)+3(0−5)=−24+40−15=1.

To ensure no two women sit together, we first arrange the 4 men in a circle. The number of ways to do this is $(4-1)! = 3! = 6$(4−1)!=3!=6 ways. This creates exactly 4 spaces between the 4 men. We then place the 4 women in these 4 linear spaces, which can be done in $4! = 24$4!=24 ways. Total arrangements $= 6 \times 24 = 144$=6×24=144.

The general term in the expansion of $(a+b)^n$(a+b)n is $\binom{n}{r} a^{n-r} b^r$(rn​)an−rbr. For the $x^2y^2$x2y2 term, we let $r=2$r=2. The term is $\binom{4}{2} (x)^2 (-2y)^2$(24​)(x)2(−2y)2. $\binom{4}{2} = 6$(24​)=6, and $(-2y)^2 = 4y^2$(−2y)2=4y2. The term is $6 \times x^2 \times 4y^2 = 24x^2y^2$6×x2×4y2=24x2y2. The coefficient is 24.

Since one particular person must be on the committee, we only need to select the remaining 2 members from the remaining 6 people. This is a combination problem: $\binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15$(26​)=2!(6−2)!6!​=2×16×5​=15 ways.

The sum to infinity is $S_\infty = \frac{a}{1-r} = 8 \Rightarrow a = 8(1-r)$S∞​=1−ra​=8⇒a=8(1−r). The sum of the first two terms is $a + ar = a(1+r) = 6$a+ar=a(1+r)=6. Substitute $a$a: $8(1-r)(1+r) = 6 \Rightarrow 8(1-r^2) = 6 \Rightarrow 1-r^2 = \frac{6}{8} = \frac{3}{4}$8(1−r)(1+r)=6⇒8(1−r2)=6⇒1−r2=86​=43​. Solving for $r^2$r2 gives $r^2 = \frac{1}{4}$r2=41​, so $r = \pm \frac{1}{2}$r=±21​. The positive value is $\frac{1}{2}$21​.

For a $2 \times 2$2×2 matrix $P = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$P=(ac​bd​), its inverse is $P^{-1} = \frac{1}{ad-bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$P−1=ad−bc1​(d−c​−ba​). The determinant is $(2)(3) - (1)(5) = 6 - 5 = 1$(2)(3)−(1)(5)=6−5=1. Swapping elements on the main diagonal and negating the others yields $\begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix}$(3−5​−12​).

The sum of the coefficients in the $n$n-th row of Pascal's triangle is given by $2^n$2n. For the 4th row, $n = 4$n=4. Thus, the sum is $2^4 = 16$24=16. Explicitly, the 4th row is 1, 4, 6, 4, 1, which sums to 16.

The formula for the sum of the first $n$n terms of an AP is $S_n = \frac{n}{2}[2a + (n-1)d]$Sn​=2n​[2a+(n−1)d]. Given $S_{10} = 185$S10​=185, $n = 10$n=10, and $a = 5$a=5: $185 = \frac{10}{2}[2(5) + 9d] \Rightarrow 185 = 5(10 + 9d)$185=210​[2(5)+9d]⇒185=5(10+9d). Dividing by 5 gives $37 = 10 + 9d \Rightarrow 27 = 9d \Rightarrow d = 3$37=10+9d⇒27=9d⇒d=3.

We need to choose 2 boys from 5, which can be done in $\binom{5}{2} = 10$(25​)=10 ways. We also need to choose 2 girls from 4, which can be done in $\binom{4}{2} = 6$(24​)=6 ways. By the multiplication principle, the total number of teams is $10 \times 6 = 60$10×6=60.