Sequences and Series Test 43

Total number of wrong answers $$= 2^{n-1}+2^{n-2}+...+2^{2}+2+1$$
$$=2^{n-1}+2^{n-2}+...+2^{4}+2^{3}+2^{2}+2+1$$
$$\Rightarrow 2^{n}-1=2043$$  [Using formula for sum of G.P]
$$ \Rightarrow 2^{n}=2048$$
$$ \Rightarrow 2^{n}=2^{11}$$
$$\Rightarrow n = 11$$
Hence, option 'B' is correct.

We know that, $$ { { n }_{ C } }_{ r }=\dfrac { n! }{ r!(n-r)! } $$

Given, $$ \dfrac { { { 19 }_{ C } }_{ r } }{ { { 19 }_{ C } }_{ r-1 } } =\quad \dfrac { 2 }{ 3 } $$

$$ =>\dfrac { \dfrac { 19! }{ r!(19-r)! }  }{ \dfrac { 19! }{ (r-1)!(19-r+1)! }  } =\quad \dfrac { 2 }{ 3 } $$

$$ => \dfrac { (r-1)!(20-r)! }{ r!(19-r)! } =\quad \dfrac { 2 }{ 3 } $$

$$ =\dfrac { (r-1)!\times (20-r) \times (19-r)! }{ r\quad \times (r-1)!\times (19-r)! } =\quad \dfrac { 2 }{ 3 } $$
$$ => \dfrac { 20-r }{ r\quad  } =\quad \dfrac { 2 }{ 3 } $$
$$ => r = 12 $$

So,
$$ { { 14 }_{ C } }_{ 12 }=\dfrac { 14! }{ 12!(14-12)! } =\dfrac { 14! }{ 12!(2)! } =\dfrac { 14\times 13\times 12! }{ 12!\times 2 } =\quad 7\times 13=91 $$

We know that $$ { { n }_{ C } }_{ r }=  { { n }_{ C } }_{n- r }$$

Given, $$ { { n }_{ C } }_{3}= { { n }_{ C } }_{13} $$
$$ => r = 3 $$ and $$ n -r = 13 $$
$$ => n - 3 = 13 $$
$$ => n = 16 $$

Also, $$ { { n }_{ C } }_{ r }=\dfrac { n! }{ r!(n-r)! }  $$
So, $$ { { 20 }_{ C } }_{ 16 }=\dfrac { 20! }{ 16!(20-16)! } = \dfrac { 20! }{ 16! \times 4! }  = \dfrac { 20 \times 19 \times 18 \times 17 \times 16! }{ 16! \times 4 \times 3 \times 2 } = 4845   $$

Here,
$$^nP_3-^nC_3 > 100$$

or $$\dfrac {n!}{(n-3)!}-\dfrac {n!}{3!(n-3)!} > 100$$

or $$\dfrac {5}{6}n(n-1)(n-2) > 100$$

or $$n(n-1)(n-2) > 100$$

or $$n(n-1)(n-2) > 6\times 5\times 4$$

or $$n=7, 8, .....$$

$$7$$ Courses in morning 

$$  {\textbf{Step 1: Find m}} $$

                $$  {\text{For m,}} $$

                $$  {\text{First we select any 5 digits from 0,1,2,}}...{\text{,9}} $$

                $$  {\text{Number of ways = }}{}^{10}{{\text{C}}_5} $$

                $$  {\text{Now after selection there is only 1 way to arrange these selected digits, i}}{\text{.e}}{\text{., in descending order}}{\text{.}} $$

                $$  {\text{Therefore m = }}{}^{10}{{\text{C}}_5}\times{\text{  1 = }}{}^{10}{{\text{C}}_5} $$

$$  {\textbf{Step 2: Find n}} $$

                $$  {\text{For n,First we select any 5 digits from 1,2,}}...{\text{,9}} $$

                $$  {\text{We can't select zero as  first digit because then the number won't be a 5 - digit number}}{\text{.}} $$

                $$  {\text{Therefore number of ways  = }}{}^9{{\text{C}}_5} $$

                $$   \Rightarrow {\text{n = }}{}^9{{\text{C}}_5}\times{\text{  1 = }}{}^9{{\text{C}}_5} $$

                $$   \Rightarrow {\text{m - n = }}{}^{10}{{\text{C}}_5}{\text{ - }}{}^9{{\text{C}}_5} $$ 

                $$  {\text{We know that,}}{}^n{{\text{C}}_r}{\text{ + }}{}^n{{\text{C}}_{r - 1}}{\text{ = }}{}^{n + 1}{{\text{C}}_r} $$

                $$   \Rightarrow {}^{n + 1}{{\text{C}}_r}{\text{ - }}{}^n{{\text{C}}_r}{\text{ = }}{}^n{{\text{C}}_{r - 1}} $$

                $$   \Rightarrow {}^{10}{{\text{C}}_5}{\text{ - }}{}^9{{\text{C}}_5}{\text{ = }}{}^9{{\text{C}}_4} $$

                $$  {\text{Hence, m - n = }}{}^9{{\text{C}}_4} $$

$$  {\textbf{Hence, the correct answer is option A}} $$

 

There are only a few ways you can make a product have a units digit of $$9$$. Either both numbers you’re multiplying have to be $$3$$, or one has to be $$1$$ and one has to be $$9$$. Since we’re dealing with consecutive integers, $$j$$ and $$n$$ can’t both end in $$3$$, so they’re going to have to end in $$1$$ and $$9$$. 

It’s important to remember that although we only really care about the units digits in this problem, we’re dealing with numbers that might (or in fact, must) be $$2$$ or more digits. That’s why you can have $$k's$$ units digit be $$0$$. 

Say $$j= 19$$, $$k= 20$$, and $$n= 21$$. Then $$jn = (19)(21) = 399$$ (units digit is $$9$$), and the units digit of $$k = 0$$.

Hence option A is correct.