Algebra: permutation and combination; binomial theorem

Prove by mathematical induction: $1^2+2^2+3^2+\cdots+n^2=\dfrac{n(n+1)(2n+1)}{6}$ for all $n\in\mathbb N$.

Let $P(n)$ be this statement.

Step 1 (n = 1): L.H.S. $=1^2=1$, R.H.S. $=\dfrac{1\cdot2\cdot3}{6}=1$. So $P(1)$ is true.

Step 2 (hypothesis): Assume $P(k)$ is true: $1^2+2^2+\cdots+k^2=\frac{k(k+1)(2k+1)}{6}.$

Step 3 (P(k+1)): Adding $(k+1)^2$ to both sides, $1^2+\cdots+k^2+(k+1)^2=\frac{k(k+1)(2k+1)}{6}+(k+1)^2=\frac{(k+1)\,[\,k(2k+1)+6(k+1)\,]}{6}$ $=\frac{(k+1)(2k^2+7k+6)}{6}=\frac{(k+1)(k+2)(2k+3)}{6}=\frac{(k+1)\{(k+1)+1\}\{2(k+1)+1\}}{6}.$ This is $P(k+1)$.

Conclusion: $P(1)$ is true and $P(k)\Rightarrow P(k+1)$, so by the principle of mathematical induction $P(n)$ is true for all $n\in\mathbb N$. $\blacksquare$

(a) DAUGHTER has 8 different letters: vowels A, U, E (3) and consonants D, G, H, T, R (5).

Treat the 3 vowels as one block. Then there are $5+1=6$ units, arranged in $6!$ ways, and the vowels can be arranged within the block in $3!$ ways.

$\text{Number of words}=6!\times3!=720\times6=4320.$

(b) Cube roots of unity: let $x^3=1$. Write $1=\cos 2k\pi+i\sin 2k\pi$. By De Moivre's theorem $x=\cos\frac{2k\pi}{3}+i\sin\frac{2k\pi}{3},\quad k=0,1,2.$

• $k=0$: $x=1$
• $k=1$: $x=-\dfrac12+\dfrac{\sqrt3}{2}i=\omega$
• $k=2$: $x=-\dfrac12-\dfrac{\sqrt3}{2}i=\omega^2$

So the cube roots are $1,\ \omega,\ \omega^2$.

(For cube roots of $-1$: $x^3+1=(x+1)(x^2-x+1)=0$ gives $x=-1,\ \dfrac{1\pm i\sqrt3}{2}$.)

(a) Total members $=6+4=10$. "At most three girls" means 0, 1, 2 or 3 girls.

Committees with more than 3 girls (all 4 girls and 1 boy): $\binom44\binom61=6$.

$\text{Required number}=\binom{10}{5}-6=252-6=246.$

Check by cases:

Total $=6+60+120+60=246$.

(b) Prove $1^2+2^2+\cdots+n^2=\dfrac{n(n+1)(2n+1)}{6}$ by induction.

Let $P(n)$ be the statement. For $n=1$: L.H.S. $=1$, R.H.S. $=\frac{1\cdot2\cdot3}{6}=1$, so $P(1)$ is true.

Assume $P(k)$: $1^2+\cdots+k^2=\dfrac{k(k+1)(2k+1)}{6}$. Then $1^2+\cdots+k^2+(k+1)^2=\frac{(k+1)[k(2k+1)+6(k+1)]}{6}=\frac{(k+1)(k+2)(2k+3)}{6}=\frac{(k+1)\{(k+1)+1\}\{2(k+1)+1\}}{6},$ which is $P(k+1)$. By the principle of mathematical induction, $P(n)$ holds for all $n\in\mathbb N$. $\blacksquare$

(c) Let a pen cost Rs $x$ and a book cost Rs $y$. $2x+3y=1100,\qquad 4x+y=700.$ $D=\begin{vmatrix}2&3\\4&1\end{vmatrix}=2-12=-10,\quad D_x=\begin{vmatrix}1100&3\\700&1\end{vmatrix}=1100-2100=-1000,\quad D_y=\begin{vmatrix}2&1100\\4&700\end{vmatrix}=1400-4400=-3000.$ $x=\frac{D_x}{D}=100,\qquad y=\frac{D_y}{D}=300.$

A pen costs Rs 100 and a book costs Rs 300. Check: $2(100)+3(300)=1100$ and $4(100)+300=700$.

(a) $P(n,r) = r! \cdot C(n,r)$

Because $C(n,r)=\dfrac{n!}{r!(n-r)!}$, multiplying both sides by $r!$ gives $r!\,C(n,r)=\dfrac{n!}{(n-r)!}=P(n,r)$.

(a) $\frac{ce - bf}{ae - bd}$

By Cramer’s rule, $x=\frac{D_x}{D}$, where $D=ae-bd$ and $D_x=ce-bf$.

(a) For $n\in\mathbb N$, $(a+b)^n=\sum_{r=0}^{n}\binom nr a^{n-r}b^r=\binom n0a^n+\binom n1a^{n-1}b+\cdots+\binom nnb^n.$

(b) The total number of permutations of $n$ distinct elements is $n!$.

(c) De Moivre's theorem: for any real $\theta$ and integer $n$, $(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta.$

(d) $1^3+2^3+\cdots+n^3=\left[\frac{n(n+1)}{2}\right]^2.$

(e) The augmented matrix is $\left(\begin{array}{cc|c}a_1&b_1&c_1\\a_2&b_2&c_2\end{array}\right).$

(a) Choose 2 boys from $x$ and 3 girls from 6: $\binom x2\binom63=120\Rightarrow\binom x2\cdot20=120\Rightarrow\frac{x(x-1)}{2}=6\Rightarrow x^2-x-12=0.$ $(x-4)(x+3)=0\Rightarrow x=4\quad(x>0).$ So there are 4 boys.

(b) The complex number is $z=(a+ib)^2=(a^2-b^2)+2abi$.

Put $a=r\cos\theta,\ b=r\sin\theta$, where $r=\sqrt{a^2+b^2}$ and $\theta=\tan^{-1}\dfrac ba$. Then $z=(r\cos\theta+ir\sin\theta)^2=r^2(\cos2\theta+i\sin2\theta)$ by De Moivre's theorem. Hence the polar form is $z=(a^2+b^2)\left[\cos2\theta+i\sin2\theta\right],\qquad \theta=\tan^{-1}\frac ba.$

a. [2]

As stated, the value of $a$ cannot be determined uniquely because $n$ and $x$ are not specified. For even $n=2m$, the middle term is $T_{m+1}=\binom{2m}{m}x^m a^m=1120,$ so $a$ depends on both $m$ and $x$. For odd $n$, there are two middle terms. Hence the question as stated has no unique value of $a$.

If the intended expansion was $(2+a)^8$, then $n=8$, so the middle term is $T_5$: $T_5=\binom{8}{4}2^4a^4=70\cdot16a^4=1120a^4.$ Given $1120a^4=1120$, we get $a^4=1$, so $a=\pm1$; if $a>0$, then $a=1$.

b. [3]

We have to prove: $1^2+2^2+3^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}.$

Let $P(n)$ be the statement.

Step 1: Base case $n=1$

$\text{LHS}=1^2=1,$ $\text{RHS}=\frac{1(1+1)(2\cdot1+1)}{6}=\frac{1\cdot2\cdot3}{6}=1.$

So $P(1)$ is true.

Step 2: Induction hypothesis

Assume $P(k)$ is true: $1^2+2^2+\cdots+k^2=\frac{k(k+1)(2k+1)}{6}.$

Step 3: Prove $P(k+1)$

$1^2+2^2+\cdots+k^2+(k+1)^2$ $=\frac{k(k+1)(2k+1)}{6}+(k+1)^2$ $=(k+1)\left[\frac{k(2k+1)}{6}+(k+1)\right]$ $=(k+1)\left[\frac{2k^2+k+6k+6}{6}\right]$ $=(k+1)\left[\frac{2k^2+7k+6}{6}\right]$ $=\frac{(k+1)(k+2)(2k+3)}{6}$ $=\frac{(k+1)\{(k+1)+1\}\{2(k+1)+1\}}{6}.$

Thus $P(k+1)$ is true whenever $P(k)$ is true. Hence, by mathematical induction, $\boxed{1^2+2^2+3^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}}.$

c. [3]

The given equations are: $x+y+z=6,$ $x-y+z=2,$ $2x+y-z=1.$

In matrix form, $AX=B,$ where

Now,

The cofactors are: $C_{11}=0,\quad C_{12}=3,\quad C_{13}=3,$ $C_{21}=2,\quad C_{22}=-3,\quad C_{23}=1,$ $C_{31}=2,\quad C_{32}=0,\quad C_{33}=-2.$

So,

Hence,

Therefore,

Thus, $\boxed{x=1,\quad y=2,\quad z=3}.$

Check: $1+2+3=6,\quad 1-2+3=2,\quad 2(1)+2-3=1.$

(c) $0$ or $1$

Since
[ P(n,r)=C(n,r)\implies \frac{n!}{(n-r)!}=\frac{n!}{r!(n-r)!} ] [ \implies 1=\frac1{r!}\implies r!=1. ] For nonnegative integer (r), (0!=1) and (1!=1), so
[ r=0 \quad \text{or} \quad r=1. ] Thus the correct answer is (c) $0$ or $1$.

(b) $\begin{pmatrix}1&0&\vdots&a\\0&1&\vdots&b\end{pmatrix}$

The final reduced form of an augmented matrix is its reduced row-echelon form, in which the coefficient part must reduce to the identity matrix $I$ (leading 1's on the diagonal, zeros elsewhere), while the last column gives the values of the unknowns, i.e. $x=a,\ y=b$. Any option that still contains zeros or non-unit pivots in the coefficient block is only an intermediate step, not the final reduced form.

(n − 1)!/2 ways (for n ≥ 3).

Reason: n different beads in a circle give (n − 1)! arrangements, and since a bracelet can be turned over (clockwise and anticlockwise orders are the same), this is halved.

Euler form of $z=re^{i\theta}$, where $r=|z|$ and $\theta=\arg z$.

For $z=i$: $r=1$ and $\theta=\dfrac\pi2$ (it lies on the positive imaginary axis). Hence $i=e^{i\pi/2}.$

The exponential series (valid for all real $x$): $e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots=\sum_{n=0}^\infty\frac{x^n}{n!}.$

Similarly, for $a>0$: $a^x=1+\frac{x\ln a}{1!}+\frac{(x\ln a)^2}{2!}+\frac{(x\ln a)^3}{3!}+\cdots.$

The expansion $(x+y)^n$ has $n+1$ terms.

• If $n$ is even, there is one middle term, the $\left(\dfrac n2+1\right)$-th term: $T_{\frac n2+1}=\binom n{n/2}x^{n/2}y^{n/2}.$
• If $n$ is odd, there are two middle terms, the $\dfrac{n+1}{2}$-th and $\dfrac{n+3}{2}$-th terms: $T_{\frac{n+1}{2}}=\binom n{\frac{n-1}2}x^{\frac{n+1}2}y^{\frac{n-1}2},\qquad T_{\frac{n+3}{2}}=\binom n{\frac{n+1}2}x^{\frac{n-1}2}y^{\frac{n+1}2}.$

The first $n$ even natural numbers are $2, 4, 6, \dots, 2n$, an arithmetic series with first term $a = 2$ and common difference $d = 2$.

$S_n = \frac{n}{2}\left[2a + (n-1)d\right] = \frac{n}{2}\left[4 + 2(n-1)\right] = \frac{n}{2}(2n + 2) = n(n+1)$

Sum of first $n$ even natural numbers $= n(n+1)$

Group 1 has 3 questions and Group 2 has (8-3=5) questions.

To attempt 5 questions with at least 2 from each group, the possible selections are:

1. 2 questions from Group 1 and 3 from Group 2
   [ \binom{3}{2}\binom{5}{3}=3\times 10=30 ]

2. 3 questions from Group 1 and 2 from Group 2
   [ \binom{3}{3}\binom{5}{2}=1\times 10=10 ]

Total number of ways: [ 30+10=40 ]

[ \boxed{40} ]

Let (x) be a cube root of unity. Then

[ x^3=1 ]

Write (1) in polar form:

[ 1=\cos(2k\pi)+i\sin(2k\pi),\quad k=0,1,2 ]

By De-Moivre’s theorem,

[ x=(1)^{1/3}=\cos\frac{2k\pi}{3}+i\sin\frac{2k\pi}{3} ]

For (k=0):

[ x=1 ]

For (k=1):

[ x=\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3} =-\frac12+\frac{\sqrt3}{2}i ]

For (k=2):

[ x=\cos\frac{4\pi}{3}+i\sin\frac{4\pi}{3} =-\frac12-\frac{\sqrt3}{2}i ]

Hence the cube roots of unity are

[ \boxed{1,\quad -\frac12+\frac{\sqrt3}{2}i,\quad -\frac12-\frac{\sqrt3}{2}i} ]

i.e. (\boxed{1,\omega,\omega^2}).

Problem (as intended): If $\sin^{-1}x+\sin^{-1}y+\sin^{-1}z=\pi$, prove that $x\sqrt{1-x^2}+y\sqrt{1-y^2}+z\sqrt{1-z^2}=2xyz.$

Proof. Let $\sin^{-1}x=A,\qquad \sin^{-1}y=B,\qquad \sin^{-1}z=C,$ so that $x=\sin A,\ y=\sin B,\ z=\sin C$ and $A+B+C=\pi .$

Now consider $\sin 2A+\sin 2B+\sin 2C .$ Using $\sin 2A+\sin 2B=2\sin(A+B)\cos(A-B)$ and $A+B=\pi-C$ (so that $\sin(A+B)=\sin C$): $\sin 2A+\sin 2B+\sin 2C=2\sin C\cos(A-B)+2\sin C\cos C$ $=2\sin C\big[\cos(A-B)+\cos C\big].$ Since $C=\pi-(A+B)$, we have $\cos C=-\cos(A+B)$. Hence $=2\sin C\big[\cos(A-B)-\cos(A+B)\big]$ $=2\sin C\,\big[2\sin A\sin B\big]\qquad(\because \cos(P-Q)-\cos(P+Q)=2\sin P\sin Q)$ $=4\sin A\sin B\sin C .$

Therefore $\sin A\cos A+\sin B\cos B+\sin C\cos C=2\sin A\sin B\sin C .$

Substituting $\sin A=x,\ \cos A=\sqrt{1-x^2}$, and similarly for $B$ and $C$: $x\sqrt{1-x^2}+y\sqrt{1-y^2}+z\sqrt{1-z^2}=2xyz .$

Hence proved. $\blacksquare$

Cramer's rule. For $a_1x+b_1y=c_1,\ a_2x+b_2y=c_2$: $D=\begin{vmatrix}a_1&b_1\\a_2&b_2\end{vmatrix},\quad D_x=\begin{vmatrix}c_1&b_1\\c_2&b_2\end{vmatrix},\quad D_y=\begin{vmatrix}a_1&c_1\\a_2&c_2\end{vmatrix}.$ If $D\ne0$, then $x=\dfrac{D_x}{D},\ y=\dfrac{D_y}{D}$. If $D=0$ and $D_x=D_y=0$ there are infinitely many solutions; if $D=0$ and $D_x$ or $D_y\ne0$ there is no solution.

Example. Solve $2x+3y=7,\ 3x+5y=13$. $D=10-9=1,\quad D_x=\begin{vmatrix}7&3\\13&5\end{vmatrix}=35-39=-4,\quad D_y=\begin{vmatrix}2&7\\3&13\end{vmatrix}=26-21=5.$ $x=-4,\qquad y=5.$

(a) $n^r$ ways

When repetition is allowed, each of the $r$ positions can be filled by any one of the $n$ things independently, so the number of arrangements is $n \times n \times \cdots \times n$ ($r$ times) $= n^r$.

(a) $\sqrt{2} e^{i\pi/4}$

For $z=1+i$, modulus $r=\sqrt{1^2+1^2}=\sqrt2$ and argument $\theta=\tan^{-1}(1)=\frac{\pi}{4}$. Hence Euler form is $z=re^{i\theta}=\sqrt2\,e^{i\pi/4}$.

$n + 1$

The expansion of $(x + y)^n$ is $(x + y)^n = \binom{n}{0}x^n + \binom{n}{1}x^{n-1}y + \cdots + \binom{n}{n}y^n.$

The binomial coefficients run from $\binom{n}{0}$ to $\binom{n}{n}$, which gives $n + 1$ terms in total.

When $n$ is even, the number of terms in the expansion of $(x+y)^n$ is $n+1$ (odd), so there is a single middle term. It is the $\left(\dfrac{n}{2}+1\right)$-th term:

$\text{Middle term} = \binom{n}{\frac{n}{2}}\, x^{\frac{n}{2}}\, y^{\frac{n}{2}}$

$2^n$

Putting $x = 1$ in $(1 + x)^n$ gives the sum of all binomial coefficients: $1 + \binom{n}{1} + \binom{n}{2} + \cdots + \binom{n}{n} = (1 + 1)^n = 2^n$

$e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots+\frac{x^n}{n!}+\cdots=\sum_{n=0}^{\infty}\frac{x^n}{n!},$ valid for all real $x$.

$\sin x = x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \dfrac{x^7}{7!} + \cdots$

This is the Maclaurin series expansion of $\sin x$, valid for all real $x$.

Since $\omega$ is a non-real cube root of unity,

$1+\omega+\omega^2=0,\qquad \omega^3=1$

Now,

$\omega^3+\omega^4+\omega^5=\omega^3(1+\omega+\omega^2)$

$=1(0)=0$

Value = 0

The system can be written as

Let

Now,

The cofactor matrix is

So,

Hence,

Now,

Therefore,

Given, [ \alpha+i\beta=\frac{1}{a+ib} ]

Rationalising the denominator, [ \alpha+i\beta=\frac{a-ib}{(a+ib)(a-ib)} =\frac{a-ib}{a^2+b^2} ] So, [ \alpha=\frac{a}{a^2+b^2},\qquad \beta=-\frac{b}{a^2+b^2} ]

Now, [ \alpha^2+\beta^2 =\frac{a^2}{(a^2+b^2)^2}+\frac{b^2}{(a^2+b^2)^2} =\frac{a^2+b^2}{(a^2+b^2)^2} =\frac{1}{a^2+b^2} ]

Therefore, [ (\alpha^2+\beta^2)(a^2+b^2) =\frac{1}{a^2+b^2}\cdot (a^2+b^2)=1 ]

Hence proved.

Let $z=3+4i$. Its modulus and argument are $r=|z|=\sqrt{3^2+4^2}=5,\qquad \theta=\tan^{-1}\frac43.$ So $z=5(\cos\theta+i\sin\theta).$

By De-Moivre’s theorem, the square roots are $\sqrt z=\sqrt5\left(\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}\right).$

Now,

Hence, $\sqrt{3+4i}=\sqrt5\left(\frac{2}{\sqrt5}+i\frac{1}{\sqrt5}\right)=\pm(2+i).$

Answer: $\boxed{\pm(2+i)}$

(b) $(a*b)*c=a*(b*c)$

The associative law says the grouping of elements does not change the result. Option (a) is the commutative law and (c) is the inverse law.

(c) $6$

The roots are conjugate surds, so the other root is $3-2\sqrt5$. Sum of roots $=(3+2\sqrt5)+(3-2\sqrt5)=6$.

The statement is the binomial theorem: for any positive integer $n$, $(x+y)^n=\binom n0x^n+\binom n1x^{n-1}y+\cdots+\binom nrx^{n-r}y^r+\cdots+\binom nny^n.$

(a) The theorem is the binomial theorem (for a positive integral index).