Permutations and Combinations Test 40

Since, $$\displaystyle n = {}^m{C_2} = \frac{{m!}}{{2!\left( {m - 2} \right)!}} = \frac{{m\left( {m - 1} \right)}}{2}$$.
and $$\displaystyle {}^n{C_2} = \frac{{n!}}{{2!\left( {n - 2} \right)!}} = \frac{{n\left( {n - 1} \right)}}{2}$$
$$\displaystyle \therefore {}^n{C_2} = \frac{{\left( {\frac{{m\left( {m - 1} \right)}}{2}} \right)\left( {\frac{{m\left( {m - 1} \right)}}{2} - 1} \right)}}{2} $$
            $$\displaystyle = \left( {\frac{{m\left( {m - 1} \right)}}{4}} \right)\left( {\frac{{m\left( {m - 1} \right)}}{2} - 1} \right)$$

            $$\displaystyle =\left( {\frac{{m\left( {m - 1} \right)}}{4}} \right)\left( {\frac{{m\left( {m - 1} \right) - 2}}{2}} \right)$$

            $$\displaystyle =\left( {\frac{{m\left( {m - 1} \right)}}{4}} \right)\left( {\frac{{{m^2} - m - 2}}{2}} \right)$$

            $$\displaystyle =\left( {\frac{{m\left( {m - 1} \right)}}{4}} \right)\left( {\frac{{\left( {m - 2} \right)\left( {m + 1} \right)}}{2}} \right)$$

            $$\displaystyle ={\frac{{\left( {m + 1} \right)m\left( {m - 1} \right)\left( {m - 2} \right)}}{{2 \cdot 4}}}$$
            $$\displaystyle =3\frac{{\left( {m + 1} \right)m\left( {m - 1} \right)\left( {m - 2} \right)\left( {\left( {m - 3} \right)!} \right)}}{{\left( {1 \cdot 2 \cdot 3 \cdot 4} \right)\left( {\left( {m - 3} \right)!} \right)}}$$
            $$\displaystyle =3\left( {{}^{m + 1}{C_4}} \right)$$

Let $$\{\mathrm{x}\}=\mathrm{x}-[\mathrm{x}]$$ denote fractional part of $$\mathrm{x}   i.e     0\leq$$ $$\{\mathrm{x}\}<1$$
Now given expression is
$$\displaystyle



\frac{x}{3}-\{\frac{x}{3}\}+\frac{3x}{2}-\{\frac{3x}{2}\}+\frac{y}{2}-\{\frac{y}{2}\}+\frac{3y}{4}-\{\frac{3y}{4}\}=\frac{11x}{6}+\frac{5y}{4}$$
$$\Rightarrow  \displaystyle \{\frac{x}{3}\}+\{\frac{3x}{2}\}+\{\frac{y}{2}\}+\{\frac{3y}{4}\}=0$$
$$\therefore  \displaystyle \{\frac{x}{3}\}=\{\frac{3x}{2}\}=\{\frac{y}{2}\}=\{\frac{3y}{4}\}=0$$
$$\Rightarrow x =6,12,18,24$$;
and $$y =4,8,12,16,20,24,28$$
Hence no. of ordered pairs $$=4\times7=28$$

$$\dfrac{\:^{n+1}C_{r+1}}{\:^{n}C_{r}}=\dfrac{11}{6}$$
$$\dfrac{(n+1)!.(n-r)!.r!}{(n-r)!(r+1)!n!}=\dfrac{11}{6}$$
$$\dfrac{n+1}{r+1}=\dfrac{11}{6}$$
$$6n+6=11r+11$$
$$6n-11r=5$$ ...(i)
and
$$\dfrac{\:^{n}C_{r}}{\:^{n-1}C_{r-1}}=2$$
$$\dfrac{n!(n-r)!(r-1)!}{(n-1)!(n-r)!r!}=2$$
$$\dfrac{n}{r}=2$$
$$n=2r$$ ...(ii)
Substituting in (i), we get
$$6(2r)-11r=5$$
$$r=5$$ and $$n=10$$

$$k^{2}-3=\displaystyle \frac{{}^{n-1}C_{r}}{{}^n{C_{r+1}}}$$
$$=\displaystyle \frac{(n-1)!}{r!.(n-r-1)!}\times\frac{(r+1)!(n-r-1)!}{n!}$$
$$=\displaystyle \frac{r+1}{n}$$
$$\therefore k^{2}=\displaystyle \frac{r+1}{n}+3$$
Also $$n-1\geq r\Rightarrow n\geq r +1(r \geq 0)$$
$$k^{2}\in(3,4]\Rightarrow k\in(\sqrt{3},2]$$

Number students who gave wrong answer to at least $$1$$ question $$=2^{n-1} = E_1 + E_2 + E_3 + .... E_n$$
Number students who gave wrong answer to at least $$2$$ questions $$=2^{n-2} = E_2 + E_3 + E_4 + ... + E_n$$
Number students who gave wrong answer to at least $$3$$ questions $$=2^{n-3} = E_3 + E_4 + E_5 + ... E_n$$
:
:
Number students who gave wrong answer to at least $$n$$ questions $$=1 = E_n$$
Number of wrong answers given $$=E_1 + 2E_2 + 3E_3 + ...+nE_n$$
Adding $$n$$ equations listed above we have,

$$\dfrac { 1.3.5...(2n-1) }{ 2.4.6...(2n) } \\ \Longrightarrow \dfrac { 1.2.3.4...(2n-1).(2n) }{ { (2.4.6...(2n)) }^{ 2 } } \\ \Longrightarrow \dfrac { 2n! }{ { ((2.1)(2.2).(2.3)...(2.n)) }^{ 2 } } \\ \Longrightarrow \dfrac { 2n! }{ { \left( { 2 }^{ n }n! \right)  }^{ 2 } } \\ \Longrightarrow \dfrac { 2n! }{ { \left( { 2 }^{ 2n })(n! \right)  }^{ 2 } } $$

$$\displaystyle \sum_{r=0}^{n-1}\dfrac {^nC_r}{^nC_r+^nC_{r+1}}$$ $$=\displaystyle \sum_{r=0}^{n-1}\dfrac {1}{1+\dfrac {^nC_{r+1}}{^nC_r}}$$ $$=\displaystyle \sum_{r=0}^{n-1}\dfrac {1}{1+\dfrac {n-r}{r+1}}$$
$$=\displaystyle \sum_{r=0}^{n-1}\dfrac {r+1}{n+1}=\dfrac {1}{n+1}\displaystyle \sum_{r=0}^{n-1}(r+1)$$
$$=\displaystyle \dfrac {1}{(n+1)}[1+2+....+n]=\dfrac {n}{2}$$

Given:
 $$3\times \left( \begin{matrix} x+1 \\ 2 \end{matrix} \right) =2\times \left( \begin{matrix} x+2 \\ 2 \end{matrix} \right) \\ \Rightarrow \dfrac { \left( \begin{matrix} x+2 \\ 2 \end{matrix} \right)  }{ \left( \begin{matrix} x+1 \\ 2 \end{matrix} \right)  } =\dfrac { 3 }{ 2 } \quad \Rightarrow \dfrac { \left( x+2 \right) \left( x+1 \right)  }{ \left( x+1 \right) x } =\dfrac { 3 }{ 2 } \\ \quad \Rightarrow 3x=2x+4\quad \quad \Rightarrow x=4$$

Here the flags are placed one above the another so, the order is important => we will be using permutations
We can hoist flags $$1$$ to $$6$$,
$$=_{1}^{6}\textrm{P}+_{2}^{6}\textrm{P}+_{3}^{6}\textrm{P}+_{4}^{6}\textrm{P}+_{5}^{6}\textrm{P}+_{6}^{6}\textrm{P}=1956$$ ways
Hence, option 'A' is correct.

Starting with G there are $$4!$$
Starting with H there are $$4!$$
Starting with O there are $$4!$$
Starting with TG there are $$3!$$
Starting with TH there are $$3!$$
Starting with TOG there are $$2!$$
Starting with TOH there are $$2!$$
Starting with TOUGH there are $$1!$$