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- Question 13 Exercise 6.2
- e total number of letters in 'Excellence' are: $n=10$, out of which $m_1=4$ are $E, m_2=2$ are $L$ and... m_3 \end{array}\right)\\&=\left(\begin{array}{c} 10 \\ 4,2,2 \end{array}\right) \\ & =\dfrac{10 !}{4 ! \cdot 2 ! \cdot 2 !}\\ &=37,800 \end{align} Begin ... e total number of letters in 'Excellence' are: $n=10$, out of which $m_1=4$ are $E$, $m_2=2$ are $L
- Question 11 Exercise 6.2
- tion 11===== How many numbers each lying between $10$ and $1000$ can be formed with digits $2.3,4,0,8,9$ using only once? ====Solution==== We will form numbers greater than $10$ and less than $1000$. So some number will consist just two digits, and some will contain three digi
- Question 5 Exercise 6.4
- n. ====Solution==== Total number of persons $=6+4=10$. Total number of ways to select $5$ out of these $10$ are: \begin{align}{ }^{10)} C_5 &=\dfrac{10 !}{(10-5) ! 5 !}\\ &=252\\ n(S)&=252\end{align} When $3$ men and $2$ women. By mu
- Question 7 Exercise 6.4
- 6$\}$, then from sample space, we see that $n(B)=10$. Thus the probability of getting a number less than 6 is: $$P(B)=\dfrac{n(B)}{n(S)}=\dfrac{10}{36}=\dfrac{5}{18}$$ =====Question 7(iii)===== T... nd the probability of getting a sum greater than $10.$ ====Solution==== The sample space rolling a pa... ts are $$n(S)=6 \times 6=36$$ A sum greater than $10$ Let $D=\{$ a sum greater than 10$\}$, then fro
- Question 1 and 2 Exercise 6.1
- an. =====Question 1(i)===== Evaluate the $\dfrac{10 !}{3 ! .3 ! \cdot 4 !}$ ====Solution==== \begin{align}\dfrac{10 !}{3 ! \cdot 3 ! \cdot 4 !}&=\dfrac{10.9 .8 \cdot 7 \cdot 6 \cdot 5.4 !}{3 ! \cdot 3 ! \cdot 4 !}\\ &=\dfrac{10.9 .8 .7 .5}{3.2 .1}\\ &=4200 \end{align} =====Qu
- Question 10 Exercise 6.5
- ====== Question 10 Exercise 6.5 ====== Solutions of Question 10 of Exercise 6.5 of Unit 06: Permutation, Combination and... KPTB or KPTBB) Peshawar, Pakistan. =====Question 10==== A basket contains $20$ apples and $10$ oranges out of which $5$ apples and $3$ oranges are defect
- Question 12 Exercise 6.2
- of letters in $\mathrm{BOOK}$ KEEPER are ten. $n=10$, out of which two are $\mathrm{O}$, so $m_1=2$,... 2, m_3 \end{array}\right)&=\left(\begin{array}{c} 10 \\ 2,3.2 \end{array}\right) \\ & =\dfrac{10 !}{2 ! \cdot 3 ! \cdot 2 !}\\ &=151,200 \end{align} ====... BBOTABAD\\ Total number of letters are ten, so $n=10$ out of which three are $\mathrm{A}$, so $m_1=3$
- Question 7 and 8 Exercise 6.3
- uestion 8===== A student is to answer $7$ out of $10$ questions in an examination. How many choices ha... Option to choose different $7$ questions out of $10$ to answer are: $${ }^{10} C_7=\dfrac{10 !}{(10-7) ! 7 !}=120$$ If he must answer the first three, then the remaining question
- Question 9 Exercise 6.3
- (7-3) ! 3 !} \cdot \dfrac{6 !}{(6-5) ! 5 !}\\\ &=210\end{align} If committee contain $4$ men then it w... }{(7-6) ! 6 !} \cdot \dfrac{6 !}{(6-2) ! 2 !}\\ &=105\end{align} If committee contain $7$ men then it ... ees that will contain at least $2$ men are: $$21+210+525+420+105+6=1,287$$ =====Question 9(iii)===== An $8$-persons committee is to be formed from a grou
- Question 10 Exercise 6.2
- ====== Question 10 Exercise 6.2 ====== Solutions of Question 10 of Exercise 6.2 of Unit 06: Permutation, Combination and... KPTB or KPTBB) Peshawar, Pakistan. =====Question 10(i)===== In how many ways can five students be sea... .4 .3 !}{3 !}\\ &=1680\end{align} =====Question 10(ii)===== In how many ways can five students be se
- Question 1 Exercise 6.3
- {(n-3) ! 3 \cdot 2 !} \\ & \Rightarrow n^2(n^2-1)=10 n(n-1)(n-2) \\ & \Rightarrow n^2(n-1)(n+1)=10 n(n-1)(n-2) \\ & \Rightarrow n(n+1)=10(n-2) \\ & \Rightarrow n^2+n-10 n+20=0\\ & \Rightarrow n^2-9 n+20=0 \\ & \Rightarrow n^2-4 n-5 n+20=0
- Question 9 & 10 Review Exercise 6
- ====== Question 9 & 10 Review Exercise 6 ====== Solutions of Question 9 & 10 of Review Exercise 6 of Unit 06: Permutation, Com... 2,3$? ====Solution==== We know that $1$ million $=100,0000$. First we are computing the total number ... than $1$ million are $420-50=360$. =====Question 10===== A party of $n$ men is to be seated around a
- Question 9 Exercise 6.5
- }{5}+\dfrac{6}{7} \cdot \dfrac{1}{5} \\ & =\dfrac{10}{35}=\dfrac{2}{7} \end{align} =====Question 9(ii... align="right"><btn type="success">[[math-11-kpk:sol:unit06:ex6-5-p7|Question 10 >]]</btn></text>
- Question 3 & 4 Review Exercise 6
- ow 51-r&=\dfrac{30800}{3080} \\ \Rightarrow r&=51-10=41\end{align} =====Question 4===== In how many d... 12 !}{4 ! 3 ! 5 !} \\ & =\dfrac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 !}{4 ! 3
- Question 7 & 8 Review Exercise 6
- hese two digits, then the remaining digits are: $$10-2=8$$ We have to fill the remaining four places w... n="right"><btn type="success">[[math-11-kpk:sol:unit06:Re-ex6-p6|Question 9 & 10 >]]</btn></text>