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- Question 13 Exercise 6.2
- of permutation of word "Excellence." How many of these permutations begin with $\mathrm{E}$ ? ====Soluti... of permutation of word "Excellence." How many of these permutations begin with $\mathrm{E}$ and end with... of permutation of word "Excellence." How many of these permutations begin with $\mathrm{E}$ and end with... of permutation of word "Excellence." How many of these permutations do not begin with $\mathrm{E}$ ? ===
- Question 5 and 6 Exercise 6.2
- er of possible arrangements to fill $7$ places by these $7$ letters are: \begin{align}^7 P_7&=\dfrac{7 !}... ,9$ ? (Repetition not being allowed). How many of these are even? ====Solution==== We have to fill four places with these five digits $2,4,5,7,9$, so that repetition is ... m{n} . \mathrm{m}$ ber that can $e$ formed though these 5 digits are: $$=5.4 .3 .2=120\quad \text{or}$$
- Question 10 Exercise 6.2
- nts are five so, $r=5$. The total number of ways these five students can be seated are: \begin{align}^8 ... wo students insist to sit next to each other then these two students will be handled as a single students... nts are five so, $r=5$. The total number of ways these five students can be seated are: \begin{align}^8 ... to each other, then the total number ways sitting these students in a row are: \begin{align}^7 P_4&= \dfr
- Question 9 & 10 Review Exercise 6
- are computing the total number of ways arranging these digits using repeated permutation as: $$=\dfrac{7... 1)$ ! If two persons sit together, we shall deal these two as one man, then in this case total will be $(n - 1)$. and the total number of ways these $(n-1)$ can sit around table are: $(n-2) !$ the ... e of counting the total number of ways of sitting these $n$ people in which two men want to sit together
- Question 5 Exercise 6.4
- +4=10$. Total number of ways to select $5$ out of these $10$ are: \begin{align}{ }^{10)} C_5 &=\dfrac{10 ... +4=10$. Total number of ways to select $5$ out of these $10$ are: \begin{align}{ }^{10} C_5 &=\dfrac{10 !
- Question 9 Exercise 6.5
- oes not depend on the other sclection, therefore these two events are independent. Thus \begin{align}P(\... but Ajmal not that is $P(\bar{A} \cup B)$. Also these both are independent events. Thus the probabilit
- Question 7 & 8 Review Exercise 6
- It means have to fill the first two places with these two digits, then the remaining digits are: $$10-2... $$ We have to fill the remaining four places with these $8$ digits, so the event $E_1$ occurs with $m_1=
- Question 11 Exercise 6.2
- i) Numbers greater than $10$ but less than $100$ These numbers will consist just two digits ten digit an
- Question 14 and 15 Exercise 6.2
- ithout any condition is $6 !$ Now, let us assume these two particular people ALWAYS sit together and let
- Question 5 and 6 Exercise 6.3
- r is "No." If there were a fourth point in one of these triangles, it would lie on a side with two of t
- Question 5 & 6 Review Exercise 6
- his case the total number of ways to give seat to these six people are: $$4 ! \cdot 2 !=48$$ Because the
- Question 11 Review Exercise 6
- {1}{4}\\ P( Green )&=\dfrac{1}{4}\end{align} Also these two are mutually exclusive events. Therefore $P(