Search
You can find the results of your search below.
Fulltext results:
- Exercise 6.2 (Solutions)
- }^n P_{n-2}$\\ [[math-11-nbf:sol:unit06:ex6-2-p1|Solution Question 1]] **Question 2.** Find $n$, if:\\ (i)... -1}=22: 7$ \\ [[math-11-nbf:sol:unit06:ex6-2-p2|Solution: Question 2 ]] **Question 3.** Find $r$, if:\\ (... +6}=1: 30800$\\ [[math-11-nbf:sol:unit06:ex6-2-p3|Solution: Question 3 ]] **Question 4.** How many 3 -digi... not repeated?\\ [[math-11-nbf:sol:unit06:ex6-2-p4|Solution: Question 4 & 5 ]] **Question 5.** How many 7 -d
- Exercise 6.3 (Solutions)
- ^{n+1} C_{r}$\\ [[math-11-nbf:sol:unit06:ex6-3-p1|Solution: Question 1(i-v)]] **Question 1(vi-x).** Prove t... ible by $k$ !\\ [[math-11-nbf:sol:unit06:ex6-3-p2|Solution: Question 1(vi-x) ]] **Question 2.** Find $n$, i... C_{2}=12: 1$\\ [[math-11-nbf:sol:unit06:ex6-3-p3|Solution: Question 2 ]] **Question 3.** Find $r$, if:\\ (... {r}-1}=11: 5$\\ [[math-11-nbf:sol:unit06:ex6-3-p4|Solution: Question 3 ]] **Question 4.** Find $n$ and $r$,
- Exercise 6.1 (Solutions)
- !}{(6!)^2}$ \\ [[math-11-nbf:sol:unit06:ex6-1-p1|Solution: Question 1]] **Question 2.** Write the followin... -1)}{n(n-4)}$\\ [[math-11-nbf:sol:unit06:ex6-1-p2|Solution: Question 2 ]] **Question 3.** Prove the followi... n^{2}-3 n+2$ \\ [[math-11-nbf:sol:unit06:ex6-1-p3|Solution: Question 3 & 4]] **Question 4.** Show that:\\ (... dots(2 n-1))$\\ [[math-11-nbf:sol:unit06:ex6-1-p3|Solution: Question 3 & 4]] **Question 5.** Find the value
- Question 2, Exercise 6.2
- )===== Find $n$, if: $\quad ^nP_4=20\, ^nP_2$ ** Solution. ** \begin{align*} \dfrac{m}{(n-4)!}&=20 \cdot \... = Find $n$, if: $\quad ^{2n}P_3=100 \, ^nP_2$ ** Solution. ** \begin{align*} \dfrac{(2 n)!}{(2 n-3)!}&=100... ind $n$, if: $\quad16\, ^nP_3=13\, ^{n+1}P_3$ ** Solution. ** \begin{align*}16 \dfrac{n!}{(n-3)!}&=13 \dfr... )===== Find $n$, if: $\quad ^nP_5=20, ^nP_3$ ** Solution. ** \begin{align*}{ }^{n} P_{5}&=20{ }^{n} P_{3}
- Question 3, Exercise 6.2
- ion 3(i)===== Find $r$, if: $^6P_{r-1}=^5P_4$ ** Solution. ** \begin{align*}{ }^{6} P_{r-1}&={ }^{5} P_{4}... (ii)===== Find $r$, if: $^{10}P_{r}=2\,^9P_r$ ** Solution. ** \begin{align*}{ }^{10} P_{r}&=2 \times{ }^{9... on 3(iii)===== Find $r$, if: $^{15}P_{r}=210$ ** Solution. ** \begin{align*}^{15}P_{r}&=210\\ \dfrac{15!}{... == Find $r$, if: $^{10}P_{r}=3\,^{10}P_{r-1}$ ** Solution. ** \begin{align*}^{10}P_{r}&=3\,^{10}P_{r-1}\\
- Question 7(i-vi), Exercise 6.1
- $\quad \dfrac{n!}{(n-2)!}=930,\quad n \geq 2$ ** Solution. ** \begin{align*} \dfrac{n!}{(n-2)!}&=930\\ \df... }=20\cdot \dfrac{n!}{(n-3)!}, \quad n \geq 5$ ** Solution. ** \begin{align*} \dfrac{n!}{(n-5)!}&=20\cdot \... == Find $n$, if $\quad (n+2)!= 60\cdot(n-1)!$ ** Solution. ** \begin{align*} (n+2)!&= 60(n-1)!\\ (n+2)!&= ... ==== Find $n$, if $\quad (n+2)!= 132\cdot n!$ ** Solution. ** \begin{align*} (n+2)!&= 132\cdot n!\\ (n+2)(
- Question 2, Exercise 6.3
- )===== Find $n$, if : $\,\, ^nC_5=\,\, ^nC_8$ ** Solution. ** \begin{align*}\dfrac{n!}{5!(n-5)!}=\dfrac{n!... === Find $n$, if : $\,\, ^nC_{15}=\,\, ^nC_7$ ** Solution. ** Since \begin{align*}{ }^{n} C_{r}&={ }^{n} C... === Find $n$, if : $\,\, ^nC_{50}=\,\, ^nC_1$ ** Solution. ** As we know \begin{align*}{ }^{n} C_{1}&={ }^... Find $n$, if : $\,\, ^{2n}C_3:^nC_3=\,\,11:1$ ** Solution. ** \begin{align*}\dfrac{(2 n)!}{(2 n-3)!}: \dfr
- Question 1, Exercise 6.1
- an. =====Question 1(i)===== Evaluate $10!$. ** Solution. ** \begin{align*} 10! &= 10 \times 9 \times 8 ... 1(ii)===== Evaluate $\dfrac{12!}{7! 3! 2!}$. ** Solution. ** \begin{align*} \dfrac{12!}{7! \, 3! \, 2!} &... n 1(iii)===== Evaluate $\dfrac{4!-2!}{3!+5!}$ ** Solution. ** \begin{align*} \dfrac{4! - 2!}{3! + 5!} &= \... 1(iv)===== Evaluate $\dfrac{(n-1)!}{(n-2)!}$. ** Solution. ** \begin{align*} \dfrac{(n-1)!}{(n-2)!} &= \df
- Question 6(i-v), Exercise 6.1
- d (2n)!=2^n(n!)[1\cdot3\cdot5 \cdots (2n-1)]$ ** Solution. ** \begin{align*} (2n)!&= (2n)(2n-1)(2n-2)(2n-3... )!(n-1)]=(n+2)!$ FIXME problem in third term ** Solution. ** \begin{align*}L.H.S.&= (n+1)[n!n+(n-1)!(2n-1)... c{n!}{r!(n-r+1)!}=\dfrac{(n+1)!}{r!(n-r+1)!}$ ** Solution. ** =====Question 6(iv)===== Prove for $n\i... quad \dfrac{n!}{r!}=n(n-1) (n-2)\cdots (r+1)$ ** Solution. ** =====Question 6(v)===== Prove for $n\i
- Question 7(vii-xi), Exercise 6.1
- ==== Find $n$, if $\quad n!=990 \cdot (n-3)!$ ** Solution. ** \begin{align*} n!&=990 (n-3)!\\ n(n-1)(n-1)... == Find $n$, if $\quad (n+1)!=6 \cdot (n-1)!$ ** Solution. ** \begin{align*} (n+1)!&=6 (n-1)!\\ (n+1)n(n-... 2n-1)!}=\dfrac{(n+3)!}{(2n+1)!}\dfrac{72}{7}$ ** Solution. ** \begin{align*} \dfrac{(n+2)!}{(2n-1)!}&=\dfr... n)!}{4!(2n-3)!}\cdot \dfrac{4!(n-4)!}{n!}=52$ ** Solution. ** \begin{align*} \dfrac{(2n)!}{4!(2n-3)!} \df
- Question 1, Exercise 6.2
- or $n \in N$: $\quad^nP_r=\dfrac{n!}{(n-r)!}$ ** Solution. ** Let us have $n$ distinct objects and we want... = Prove for $n \in N$: $\quad^nP_n=^nP_{n-1}$ ** Solution. ** \begin{align*} L.H.S&=^{n} p_{n}=\dfrac{n!}{... r $n \in N$: $\quad^nP_r=n\quad^{n-1}P_{r-1}$ ** Solution. ** \begin{align*}R. H.S &=n^{n-1} P_{r-1}\\ & =... $: $\quad^nP_r=^{n-1}P_r+r\quad^{n-1}P_{r-1}$ ** Solution. ** \begin{align*}R.H.S &={ }^{n-1} P_{r}+r^{n-1
- Question 1(i-v), Exercise 6.3
- r $n\in N$: $\quad^nC_r=\dfrac{n!}{r!(n-r)!}$ ** Solution. ** Let us have $n$ distinct objects and we wan... $n\cdot^{n-1}C_{r-1}=(n-r+1)\quad ^nC_{r-1}$ ** Solution. ** \begin{align*}L.H.S &=n\cdot^{n-1}C_{r-1} \\... ve for $n\in N$: $r\, ^nC_r=(n-r+1)^nC_{r-1}$ ** Solution. ** \begin{align*} L.H.S &=r \cdot \dfrac{n!}{r!... in N$: $\quad^{n-1}C_{r-1}+^{n-1}C_{r}=^nC_r$ ** Solution. ** \begin{align*} L.H.S &=\dfrac{(n-1)!}{(r-1)!
- Question 1(vi-x), Exercise 6.3
- ^{2n}C_n=\dfrac{2^n[1.3.5.\cdots(2n-1)]}{n!}$ ** Solution. ** \begin{align*}L.H.S &=\quad^{2n}C_n \\ &=\df... $\quad^nC_p=^nC_q\implies p=q\,\,or\,\,p+q=n$ ** Solution. ** Let \begin{align*}{ }^{n} C_{p}&={ }^{n} C_{... $\,\,^nC_r+2^nC_{r-1}+^nC_{r-2}=\,^{n+2}C_r$ ** Solution. ** \begin{align*}L.H.S& ={ }^{n} C_{p}+2^{n} C_... in N$: $\,\,r\cdot^nC_r=\,n\,\,^{n-1}C_{r-1}$ ** Solution. ** \begin{align*} L.H.S &=r\cdot^nC_r\\ &=r \cd
- Question 7 and 8, Exercise 6.3
- can it be done if it has exactly $2$ women. ** Solution. ** (i) If there are exactly $2$ women then ther... can it be done if it has at least $2$ women. ** Solution. ** At least $2$ women means there could be more... can it be done if it has at most $2$ women? ** Solution. ** At most to women mean either $1$ or $2$ wome... $ points on circle. Find the number of lines? ** Solution. ** For a line, we need only two points so numbe
- Review Exercise (Solutions)
- alphabets?\\ [[math-11-nbf:sol:unit06:Re-ex6-p2|Solution: Question 2 & 3 ]] **Question 3.** How many $3$-... unit place?\\ [[math-11-nbf:sol:unit06:Re-ex6-p2|Solution: Question 2 & 3 ]] **Question 4.** How many $6$... repeating?\\ [[math-11-nbf:sol:unit06:Re-ex6-p3|Solution: Question 4, 5 & 6 ]] **Question 5.** The numbe... key chain.\\ [[math-11-nbf:sol:unit06:Re-ex6-p3|Solution: Question 4, 5 & 6 ]] **Question 6.** Tweleve