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- Question 3 & 4 Exercise 6.1
- alue of $n$, when $\dfrac{n(n !)}{(n-5) !}=\dfrac{12(n !)}{(n-4) !}$ ====Solution==== We are given: \begin{align}\dfrac{n(n !)}{(n-5) !}&=\dfrac{12(n !)}{(n-4) !} \\ \Rightarrow \dfrac{n}{(n-5) !}&=\dfrac{12}{(n-4)(n-5) !} \\ \Rightarrow n&=\dfrac{12}{(n-4)} \\ \Rightarrow n(n-4)&=12 \\ \rightarrow n^2-4 n-
- Question 12 Exercise 6.2
- ====== Question 12 Exercise 6.2 ====== Solutions of Question 12 of Exercise 6.2 of Unit 06: Permutation, Combination and... KPTB or KPTBB) Peshawar, Pakistan. =====Question 12(i)===== How many different word can be formed fro... ot 3 !}{3 !}\\ &=6,720 \end{align} =====Question 12(ii)===== How many different word can be formed fr
- Question 5 and 6 Exercise 6.3
- )===== How many straight lines are determined by $12$ points, no three of which lie on the same straig... ne? ====Solution==== Total points are eight so $n=12$. Every pair of points determines a line, so that gives ${ }^{12} C_2=66$ lines. The question now is whether we ... 5(ii)===== How many triangles are determined by $12$ points, no three of which lie on the same straig
- Question 7 Exercise 6.4
- $$P(A)=\dfrac{n(A)}{n(S)}=\dfrac{3}{36}=\dfrac{1}{12}$$ =====Question 7(ii)===== Two dice are thrown ... $P(C)=\dfrac{n(C)}{n(S)}=\dfrac{15}{36}=\dfrac{5}{12}$$ =====Question 7(iv)===== Two dice are thrown ... $$P(D)=\dfrac{n(D)}{n(S)}=\dfrac{3}{36}=\dfrac{1}{12}$$ =====Question 7(v)===== Two dice are thrown ... },$ then we see from the sample space that $n(K)=12$. Thus the probability is: $$P(K)=\dfrac{n(K)}{n
- Question 3 & 4 Review Exercise 6
- number of letter in this word are twelve, so $n=12$ out of which four are $x$ so, $m_1=4$, three ar... _2, m_3 \end{array}\right)=\left(\begin{array}{c} 12 \\ 4,3,5 \end{array}\right)&=\dfrac{12 !}{4 ! 3 ! 5 !} \\ & =\dfrac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 !}{4 ! 3 !
- Question 3 Exercise 6.3
- !}{(2 n-3) ! 3 !} \times \dfrac{(n-2) ! 2 !}{n !}=12 \\ & \Rightarrow \dfrac{2 n(2 n-1)(2 n-2)(2 n-3) ... 3) ! 3 !}\times\dfrac{(n-2) ! 2 !}{n(n-1)(n-2) !}=12 \\ & \Rightarrow \dfrac{4 n(2 n-1)(n-1)}{3} \cdot \dfrac{1}{n(n-1)}=12 \\ & \Rightarrow 2 n-1=9 \\ & \Rightarrow 2 n=10
- Question 1 and 2 Exercise 6.1
- n} =====Question 2(ii)===== Write $2.4.6 .8 .10 .12$ in term of factorial. ====Solution==== \begin{align}2.4,6.8 .10 .12&=2 \cdot(2 \times 2) \cdot(2 \times 3) \cdot(2 \t
- Question 6 Exercise 6.4
- e total number of face cards in deck of card are $12.$ Thus the probability of getting a face card is: $$=\dfrac{12}{52}=\dfrac{3}{13}$$ ====Go To==== <text ali
- Question 1 and 2 Exercise 6.2
- \Rightarrow(n-3)(n-4)&=56 \\ \Rightarrow n^2-7 n+12-56&=0 \\ \Rightarrow n^2-7 n-44&=0 \\ \Rightarrow
- Question 5 and 6 Exercise 6.2
- e$ formed though these 5 digits are: $$=5.4 .3 .2=120\quad \text{or}$$ can be found using permutation as: $$^5 P_4=\dfrac{5 !}{5-4} !=120$$ Even Numbers Out of these for even number, t... 1 \cdot m_2 \cdot m_3 \cdot m_4=2 \cdot 3 \cdot 2=12$$ ====Go To==== <text align="left"><btn type=
- Question 11 Exercise 6.2
- er than $10$ and less than $1000$ are: $$100 + 25=125$$ ====Go To==== <text align="left"><btn typ... align="right"><btn type="success">[[math-11-kpk:sol:unit06:ex6-2-p8|Question 12 >]]</btn></text>
- Question 13 Exercise 6.2
- t)\\&=\dfrac{9 !}{3 ! \cdot 2 ! \cdot 2 !}\\ &=15,120 \end{align} =====Question 13(ii)===== Find the ... tation\, begin\, with\,} \mathrm{E}\\ &=37,800-15,120\\ &=22,680 \end{align} =====Question 13(v)====... ry">[[math-11-kpk:sol:unit06:ex6-2-p8 |< Question 12 ]]</btn></text> <text align="right"><btn type="su
- Question 14 and 15 Exercise 6.2
- frac{(n-1) !}{2}=\dfrac{(5-1) !}{2}=\dfrac{24}{2}=12 $$ =====Question 15===== In how many ways can $7
- Question 1 Exercise 6.3
- !}=\dfrac{6}{2 !} \\ & \Rightarrow \dfrac{n(n+1)}{12}=6 \\ & \Rightarrow n^2+n-72=0 \\ & \Rightarrow n
- Question 1 Review Exercise 6
- the digits $3$ and $7$ must together? * (a) $120$ * (b) $180$ * %%(c)%% $144$ * %%(d... wer</btn><collapse id="a3" collapsed="true">(a): $120$</collapse> iv. Evaluate $\dfrac{(n+2) !(n-2) !... tee of 4 people will be selected from 8 girls and 12 boys in a class. How many different selections ar... * %%(a)%% $18$ * (b) $27$ * %%(c)%% $512$ * (d) $81$ \\ <btn type="link" collapse="a7