Permutations and Combinations Test 43

Given, number of vowels =number of consonants
So, there will be 2 cases (i) no repetition(number of vowels =number of consonants=2)
                                         (ii) both the vowel and consonant will be repeated twice(number of vowels =number of consonants=1)
The method that will be followed involves selection of 2 places out of the 4 places which can be done in 6 ways and then applying the above 2 cases
case (i): there will be 21 ways for secting the first consonant and 20 ways for the second and There will be 5 ways of selecting the first vowel and 4 ways of selecting the second.
that makes number of words in case (i) = $$6(21\times 20\times 5\times 4)$$
case (ii): we have to select only 1 out of the 21 consonants and 5 vowels and both of them will be repeated twice
$$\Longrightarrow $$number ofwords in case (ii) =$$6(21\times 5)$$
\therefore number of words$$= 6(21\times 20\times 5\times 4)$$+$$6(21\times 5)$$
                                              $$= 6(21\times 5)(20\times 4+1)$$
                                              $$= 210\times243$$

Expanding, we get

$${\textbf{Step 1: Consider a matrix which has 4 elements.}}$$

               $${\text{As a matrix contains four elements so, the order of a matrix can be,}}$$ $$1 \times 4$$ $${\text{or}}$$ $$4 \times 1$$ $${\text{or}}$$ $$2 \times 2.$$

               $${\text{Also in every such matrix, each element is independent and has 4 different choices}}$$ $$\left( {0, 1, 2, 3} \right)$$

               $$\therefore $$ $${\text{The number of ways to fill four places of a matrix of order 4}} \times {\text{1 by 0, 1, 2, 3}}$$

                $$ = {}^4{C_1} \times {}^4{C_1} \times {}^4{C_1} \times {}^4{C_1}$$

               $$\therefore $$ $${\text{The number of ways to fill four places of a matrix of order 4}} \times {\text{1 by 0, 1, 2, 3}}$$ $$ = 4 \times 4 \times 4 \times 4$$           

               $$\left(\mathbf {\because {}^n{C_1} = n} \right)$$

               $$\therefore $$ $${\text{The number of ways to fill four places of a matrix of order 4}} \times {\text{1 by 0, 1, 2, 3}}$$ 

               $$ = {4^4} \ldots \left( 1 \right)$$

$${\textbf{Step 2: Find total number of matrices that can be formed.}}$$

               $${\text{From equation }}\left( 1 \right)$$ $${\text{we can say that, number of matrices in each order is }}{{\text{4}}^4}.$$

               $${\text{Therefore,}}$$

               $${\text{The number of matrices that can be formed}}$$ $$ = {4^4} + {4^4} + {4^4}$$

               $$ \Rightarrow $$ $${\text{The number of matrices that can be formed}}$$ $$ = 3 \times {4^4}$$

$${\textbf{Hence, option (C)}}{\textbf{ 3}} \mathbf {\times {4^4}}$$ $${\textbf{is correct answer.}}$$  

Expanding we get
$$(1\:^{n}C_{1}+2\:^{n}C_{2}+...n\:^{n}C_{n})+(1\:^{1}P_{1}+2\:^{2}P_{2}+...n\:^{n}P_{n})$$
$$=\frac{d(1+x)^{n}}{dx}|_{x=1}+(1\:^{1}P_{1}+2\:^{2}P_{2}+...n\:^{n}P_{n})$$
$$=n(1+x)^{n-1}|_{x=1}+(1!+2(2!)+3(3!)+4(4!)+...n(n!))$$
$$=n2^{n-1}+(1!+2(2!)+3(3!)+4(4!)+...n(n!))$$
Now consider
$$S_{n}=1!+2(2!)+3(3!)+4(4!)+...n(n!)$$
$$S_{1}=1$$
$$=2!-1$$
$$S_{2}=5$$
$$=3!-1$$
$$S_{3}=1+4+18$$
$$=23$$
$$=4!-1$$
Hence
$$S_{n}=(n+1)!-1$$
Hence the final answer is
$$n2^{n-1}+(n+1)!-1$$

Total number of signals can be made by using at a time one or more but not larger than five flags.
Now number of signals when r flags are used at a time from 5 flags is equal to the number of arrangement of 5 taking r at a time i.e $$^{5}P_{r}(r=1,2,...5)$$
$$\therefore$$ Required ways
$$^{5}P_{1}+^{5}P_{2}+....+^{5}P_{5}$$
$$=5+20+60+120+120$$
$$=325$$

Simplifying, we get
$$\:^{47}C_{4}+\:^{46}C_{3}+\:^{47}C_{3}+\:^{48}C_{3}+\:^{49}C_{3}+\:^{50}C_{3}+\:^{51}C_{3}+\:^{52}C_{3}$$

The greatest coefficient will be 
$$\:^{200}C_{100}$$
$$=\dfrac{200!}{100!.100!}$$
Now the exponent of 7 in 200! will be 
$$[\dfrac{200}{7}]+[\dfrac{200}{49}]$$
$$=[28.57]+[4.08]$$
$$=28+4$$
$$=32$$
The exponent of 7 in 100! will be 
$$[\dfrac{100}{7}]+[\dfrac{100}{49}]$$
$$=[14.28]+[2.04]$$
$$=14+2$$
$$=16$$
Hence the power of 7 in $$(100!)^{2}$$ will be 
$$16+16$$
$$=32$$
Hence the powers of seven in the numerator and denominator will be equal.
Hence answer is $$0$$.

We know: $${ (1+x) }^{ n }=_{ 0 }^{ n }{ C }+_{ 1 }^{ n }{ C }x+_{ 2 }^{ n }{ C }{ x }^{ 2 }+...+_{ n }^{ n }{ C }{ x }^{ n }$$
Integrating from 0 to 1: $$\dfrac { { (1+x) }^{ n+1 } }{ n+1 }-\dfrac { 1 }{ n+1 } =_{ 0 }^{ n }{ Cx+ }_{ 1 }^{ n }{ C\dfrac { { x }^{ 2 } }{ 2 }  }+...+_{ n }^{ n }{ C\dfrac { { x }^{ n+1 } }{ n+1 }  }$$
Put $$x=1$$ above: $$\dfrac { { 2 }^{ n+1 } }{ n+1 } =_{ 0 }^{ n }{ C+ }\dfrac { _{ 1 }^{ n }{ C } }{ 2 } +...+\dfrac { _{ n }^{ n }{ C } }{ n+1 } $$
Put $$x=-1$$ above: $$-\dfrac { 1 }{ n+1 } =-_{ 0 }^{ n }{ C+ }\dfrac { _{ 1 }^{ n }{ C } }{ 2 } -\dfrac { _{ 2 }^{ n }{ C } }{ 3 } ...+{ (-1) }^{ n }\dfrac { _{ n }^{ n }{ C } }{ n+1 } $$
Subtracting the above 2 results and using $$n = 20$$:
$$2(^{20}C_0+\dfrac { _{ 2 }^{ 20 }{ C } }{ 3 } +...+\dfrac { _{ 20 }^{ 20 }{ C } }{ 21 } )=\dfrac { { 2 }^{ 21 } }{ 21 }  $$
Hence, mean = $$\dfrac { \dfrac { 1 }{ 2 } \times \dfrac { { 2 }^{ 21 } }{ 21 } }{ 11 } =\dfrac { { 2 }^{ 20 } }{ 3\times77 } $$ (since there are 11 terms)
Hence, (c) is correct.

$$\displaystyle \frac{^{n}C_{r}+3^{n}C_{r+1}+3^{n}C_{r+2}+^{n}C_{r+3}}{^{n}C_{r}+4^{n}C_{r+1}+6^{n}C_{r+2}+4^{n}C_{r+3}+^{n}C_{r+4}}$$
$$\displaystyle =\frac{r+k}{n+k}$$
$$\displaystyle \Rightarrow \frac{r+4}{n+4}=\frac{r+k}{n+k}$$
$$\Rightarrow k=4$$

$$^{n+4}C_{r}$$-$$^{n}C_{r}$$-3.$$^{n}C_{r-1}$$-3.$$^{n}C_{r-2}$$-$$^{n}C_{r-3}$$
$$\Longrightarrow $$$$^{n+4}C_{r}$$-(($$^{n}C_{r}$$+$$^{n}C_{r-1}$$)+2.($$^{n}C_{r-1}$$+$$^{n}C_{r-2}$$)+($$^{n}C_{r-2}$$+$$^{n}C_{r-3})$$
$$\Longrightarrow $$$$^{n+4}C_{r}$$-($$^{n+1}C_{r}$$+2.$$^{n+1}C_{r-1}$$+$$^{n+1}C_{r-2}$$)
$$\Longrightarrow $$$$^{n+4}C_{r}$$-(($$^{n+1}C_{r}$$+$$^{n+1}C_{r-1}$$)+($$^{n+1}C_{r-1}$$+$$^{n+1}C_{r-2}$$))
$$\Longrightarrow $$$$^{n+4}C_{r}$$-($$^{n+2}C_{r}$$+$$^{n+2}C_{r-1}$$)
$$\Longrightarrow $$$$^{n+3}C_{r}$$+$$^{n+3}C_{r-1}$$-$$^{n+3}C_{r}$$
$$\Longrightarrow $$$$^{n+3}C_{r-1}$$