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- Question 2 & 3, Exercise 1.1 @math-11-kpk:sol:unit01
- uestion 2 & 3 of Exercise 1.1 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Gr... n} GOOD =====Question 3(i)===== Add the complex numbers $3\left( 1+2i \right),-2\left( 1-3i \right)$. ===... align} =====Question 3(ii)===== Add the complex numbers $\dfrac{1}{2}-\dfrac{2}{3}i,\dfrac{1}{4}-\dfrac{1... ign} =====Question 3(iii)===== Add the complex numbers $\left( \sqrt{2},1 \right),\left( 1,\sqrt{2} \rig
- Question 1, Exercise 1.1 @math-11-kpk:sol:unit01
- of Question 1 of Exercise 1.1 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Gr
- Question 3 & 4 Exercise 4.3 @math-11-kpk:sol:unit04
- kistan. =====Question 3===== Find sum of all the numbers divisible by $5$ from $25$ to $350$. GOOD ====Solution==== The numbers divisible by $5$ from $25$ tò $350$ are\\ $$25,30... end{align} =====Question 4===== The sum of three numbers in an arithmetic sequence is $36$ and the sum of ... d them. ====Solution==== Let us suppose the three numbers are $a-d, a, a+d$\\. then by first condition the
- Question 3 and 4 Exercise 4.2 @math-11-kpk:sol:unit04
- eshawar, Pakistan. =====Question 3===== Find the numbers of terms in arithmetic progression $6,9,12, \ldot
- Question 1 Review Exercise 7 @math-11-kpk:sol:unit07
- ): $2520$</collapse> ii. How many two digits odd numbers can be formed form the digits $\{1,2,3,4,5,6,7\}$
- Question 14 Exercise 7.1 @math-11-kpk:sol:unit07
- $2^{2 n}-1$ is a multiple of $3$ for all natural numbers. ====Solution==== 1. For $n=1$ then $$2^{2 n}-1=2
- Question 9 & 10 Review Exercise 6 @math-11-kpk:sol:unit06
- eshawar, Pakistan. =====Question 9===== How many numbers greater than a million can be formed with the dig... should not start with $0$, therefore the total numbers that do not start with zero. It can be formed us... are: $$=\dfrac{6 !}{2 ! 3 !}=60 $$ Thus the total numbers greater than $1$ million are $420-50=360$. =====
- Question 7 & 8 Review Exercise 6 @math-11-kpk:sol:unit06
- ====Question 8===== How many six digits telephone numbers can be constructed with the digits $0,1,2,3,4,5,6
- Question 1 Review Exercise 6 @math-11-kpk:sol:unit06
- ): $2520$</collapse> ii. How many two digits odd numbers can be formed form the digits $\{1,2,3,4,5,6,7\}$
- Question 3 and 4 Exercise 6.5 @math-11-kpk:sol:unit06
- the square of an integer. ====Solution==== Total numbers written on tickets are \begin{align}S&=\{1,2,3, \... \ n(S)&=50 \end{align} Let \begin{align}A \{odd \,numbers \}&=\{1,3,5,..,29\}\\ n(A)&=15\\ \text{Let}\, B&=
- Question 7 Exercise 6.4 @math-11-kpk:sol:unit06
- . Find the probability of getting doublet of even numbers. ====Solution==== The sample space rolling a pair... oints are $$n(S)=6 \times 6=36$$ doublet of even numbers. Let \begin{align}A&=\{(2,2),(4,4),(6,6)\}\\ n(A... e that $n(F)=4$. Thus the probability of getting numbers whose product is $6$ is: $$P(F)=\dfrac{n(F)}{n(S)
- Question 13 Exercise 6.2 @math-11-kpk:sol:unit06
- $ are $L$ and $m_3=2$ are $C$. \begin{align}\text{Numbers of perinutations are}& =\left(\begin{array}{c} n
- Question 11 Exercise 6.2 @math-11-kpk:sol:unit06
- shawar, Pakistan. =====Question 11===== How many numbers each lying between $10$ and $1000$ can be formed ... 9$ using only once? ====Solution==== We will form numbers greater than $10$ and less than $1000$. So some ... e digits. Thus we split into two parts as:\\ (i) Numbers greater than $10$ but less than $100$ These numbers will consist just two digits ten digit and unit digit
- Question 7 and 8 Exercise 6.2 @math-11-kpk:sol:unit06
- n. =====Question 7(i)===== How many three digits numbers can be formed from the digits $1,2,3,4$ and 5 if ... =====Question 7(ii)===== How many three digits numbers can be formed from the digits $1,2,3,4$ and 5 if
- Question 5 and 6 Exercise 6.2 @math-11-kpk:sol:unit06
- tation as: $$^5 P_4=\dfrac{5 !}{5-4} !=120$$ Even Numbers Out of these for even number, the unit digit ha... be filled by $2$ or $4$. So, we are left with $3$ numbers. Thus Unit digit: $E_1$ occurs in $m_1=2$ Hundr... al principle of counting the total number of even numbers are: $$m_1 \cdot m_2 \cdot m_3 \cdot m_4=2 \cdot