Simplify \(\frac{^{n}P_{3}}{^{n}C_{2}} + ^{n}P_{0}\)
n - 5
n - 3
2n - 1
2n - 3
Correct answer is D
\(\frac{^{n}P_{3}}{^{n}C_{2}} + ^{n}P_{0}\)
\(\frac{^{n}P_{3}}{^{n}C_{2}} = \frac{n!}{(n - 3)!} ÷ \frac{n!}{(n - 2)! 2!}\)
\(\frac{n!}{(n - 3)!} \times \frac{(n - 2)(n - 3)! 2!}{n!} = 2n - 4\)
\(^{n}P_{0} = \frac{n!}{(n - 0)!} = 1\)
\(\frac{^{n}P_{3}}{^{n}C_{2}} + ^{n}P_{0} = 2n - 4 + 1 = 2n - 3\)
\(3, (-1, 2)\)
\(2\sqrt{2}, (2, -1)\)
\(2\sqrt{2}, (2, 1)\)
\(9, (2, 1)\)
Correct answer is C
Equation of a circle with radius r and centre (a, b).
= \((x - a)^{2} + (y - b)^{2} = r^{2}\)
Expanding, we have
\(x^{2} - 2ax + a^{2} + y^{2} - 2by + b^{2} = r^{2}\)
Comparing, with \(x^{2} + y^{2} - 4x - 2y - 3 = 0\)
\(2a = 4 \implies a = 2\)
\(2b = 2 \implies b = 1\)
\(r^{2} - a^{2} - b^{2} = 3 \implies r^{2} = 3 + 2^{2} + 1^{2} = 8\)
\(r = 2\sqrt{2}\)
\(\begin{pmatrix} 16 & 36 \end{pmatrix}\)
\(\begin{pmatrix} 8 & 18 \end{pmatrix}\)
\(\begin{pmatrix} 8 & 0 \end{pmatrix}\)
\(\begin{pmatrix} 1 & 2 \end{pmatrix}\)
Correct answer is A
\(|Q| = \begin{vmatrix} -1 & -2 \\ 3 & 2 \end{vmatrix}\)
= \(-2 - (-6) = 4\)
\(4P = 4\begin{pmatrix} 4 & 9 \end{pmatrix}\)
= \(\begin{pmatrix} 16 & 36 \end{pmatrix}\)
\(QP\)
\(P^{2}\)
\(Q^{2}P\)
\(PQ\)
Correct answer is D
A \(1 \times 2\) matrix can be multiplied by \(2 \times 2\) matrix to give a \(1 \times 2\) matrix.
How many ways can 12 people be divided into three groups of 2, 7 and 3 in that order?
7920
792
187
42
Correct answer is A
= \(\frac{12!}{2! 3! 7!} \)
\(\frac{12 \times 11 \times 10 \times 9 \times 8 \times 7!}{2 \times 3 \times 2 \times 7!}\)
= \(7,920\)