Maths M.C.Q (1 Marks)

$C(n,r) = 70$ ==>$\frac{{n!}}{{r!\,(\,n - r)!}} = 70$…..$(ii)$  $\frac{{1680}}{{r!}} = 70$, [From $(i)$ and $(ii)$]

$r! = \frac{{1680}}{{70}} = 24$==> $r = 4$

$\because $ $p(n,4) = \,1680$

$\therefore$ $n(n - 1)(n - 2)(n - 3) = 1680$ ==> $n = 8$

$8 \times 7 \times 6 \times 5 = 1680$

Now $69n + r\,! = 69 \times 8 + 4!$

$ = 552 + 24$ $= 576$.

 

The remaining $7$ persons can speak in $7\;!$ ways.

Hence, the number of ways in which all the $10$ person can speak is $^{10}{C_3}\;.\;7\;!\; = \frac{{10\;!}}{{3\;!}}. = \frac{{10\;!}}{6}.$

$\therefore $$2.{\;^n}{C_2} - 2\;.\;2n = 66$ (By hypothesis)

$ \Rightarrow $${n^2} - 5n - 66 = 0 \Rightarrow n = 11$

$\therefore $ Number of participants $ = 11\;{\rm{men}} + 2\;{\rm{women}} = 13$.

Therefore the required number is ${^7}{C_2} = 21$.

Therefore the required number of ways in which the car can be filled is $^5{C_2}{ \times ^2}{C_1} = 20$.

So thousand's place can be filled in $5$ ways.

Remaining $3$ places can be filled by the remaining $8$ digits in $^8{P_3}$ ways

. Hence required number = $5{ \times ^8}{P_3}$.

hence number of ways = $^{12}{C_4}$.

Since there are $b$ copies of each of two books, $c$ copies of each of three books and single copies of $d$ books.

Therefore the total number of arrangements is $\frac{{(a + 2b + 3c + d)\;!}}{{a\;!\;{{(b\;!)}^2}{{(c\;!)}^3}}}$.

{Since total points are $8$, but $5$ are collinear and other three are also collinear}.

Two straight lines from the first set can be chosen in $^{m + 2}{C_2}$ ways and two straight lines from the second set can be chosen in $^{m + 2}{C_2}$ ways.

Hence the total number of parallelograms formed ${ = ^{m + 2}}{C_2}\;.{\;^{m + 2}}{C_2} = {\left( {^{m + 2}{C_2}} \right)^2}$.

Similarly $11$ straight lines out of the given $37$ straight lines intersect at $B$. Therefore instead of getting $^{11}{C_2}$ points, we get only one point. Hence the number of intersection points of the lines is $^{37}{C_2}{ - ^{13}}{C_2}{ - ^{11}}{C_2} + 2 = 535$.

$(i)$ Number of lines${ = ^{18}}{C_2}{ - ^5}{C_2} + 1 = 153 - 10 + 1 = 144$

$(ii)$ Number of $\Delta s{ = ^{18}}{C_3}{ - ^5}{C_3} = 816 - 10 = 806$

The numbers of $4$ digits number divisible by $5$ are $90 \times 20 = 1800$.

Hence required number of ways are $9000 - 1800 = 7200$.

{Since there are $20$ numbers in each hundred ($1$ to $100$) divisible by $5$ and from $999$ to $9999$ there are $90$ hundreds, hence the results}.

$^{2n + 1}{C_0}{ + ^{2n + 1}}{C_1} + ..... + {\,^{2n + 1}}{C_n} = S$ (Say)

Then $2S = 2{(^{2n + 1}}{C_0} + {\,^{2n + 1}}{C_1} + ..... + {\,^{2n + 1}}{C_n})$

= ${(^{2n + 1}}{C_0} + {\,^{2n + 1}}{C_{2n + 1}}) + {(^{2n + 1}}{C_1} + {\,^{2n + 1}}{C_{2n}}) + ....$

$.... + {(^{2n + 1}}{C_n} + {\,^{2n + 1}}{C_{n + 1}})$

= $^{2n + 1}{C_0} + {\,^{2n + 1}}{C_1} + ....... + {\,^{2n + 1}}{C_{2n + 1}} = {2^{2n + 1}}$

$ \Rightarrow $ $S = {2^{2n}}$.

$ = (1 + 1)(2 + 1)(2 + 1)(1 + 1)(1 + 1) - 1 = 71$.

 The number of ways to sit $5$ members = $5!$

$6$ Places are filled by $5$ members by $^6{C_5}$ ways

 The total number of ways to sit $5$ members on $6$ seats of a bus = $^6{C_5} \times 5!$.

Required number of ways

$= 989 - \left\{ \begin{array}{l}20,\;30,\;40,................100 = 9\\101,\;102,...................200 = 19\\201,........................300 = 19\\.............................................\\.............................................\\901,..........................990 = 18\end{array} \right\}$

$ = 989 - (9 + 18 + 19 \times 8) = 810$.

Aliter : Between $10$ and $1000$, the numbers are of $2$ digits and $3$ digits.

Since repetition is allowed, so each digit can be filled in $9$ ways.

Therefore number of $2$ digit numbers $ = 9 \times 9 = 81$

and number of $3$ digit numbers $9 \times 9 \times 9 = 729$

Hence total ways $ = 81 + 729 = 810$.

This can be done in $^7{C_4}$ ways. These $5$ persons can be arranged in a line in $5\;!$ ways.

Hence the number of possible arrangements is $^7{C_4}.\;5\;!\;{ = ^7}{C_3}.\,(5\;!)$.

Then ${x_1}$ takes the value $1,\;2,\;3,\;........,9$ and ${x_2},\;{x_3},...{x_7}$ all take values $0, 1, 2, 3, ..., 9$.

If we keep ${x_1},\;{x_2},......,{x_6}$ fixed, then the sum ${x_1} + {x_2} + ..... + {x_6}$ is either even or odd.

Since ${x_7}$ takes $10$ values $0, 1, 2, ..., 9$, five of the numbers so formed will be even and $5$ odd.

Hence the required number of numbers

$ = 9\;.\;10\;.\;10.\;10\;.\;10.\;10\;.\;5 = 4500000$.

Each question can be answered in 4 different ways

Required number of ways

$={ }^8 C _5 \times(1) \times(3)^3$

$=1512$

$a b=(a !+b !)$

$\because ab \leq 99 \Rightarrow a !+ b ! \leq \frac{99}{4}$

$a, b \leq 4$

Check for $a=1,2,3,4$

$b =0,1,2,3,4$

We get  $12,32$

Sum $=12+32=44$

According to the condition $d / nd$

$\frac{10 n + d }{ d } \in I \Rightarrow \frac{10 n }{ d }+1 \in I$

$\frac{10 n }{ d } \in I \Rightarrow \frac{2 \times 5 \times n }{ d } \in I$

for smallest ' $n$ ', $2 \times 5 \times n$ is always divisible by $d \in\{1,2, \ldots, 9\}$

$n =3^2 \times 2^2 \times 7$

$n =252$

Product of digits of ' $n$ ' $=20$

Here, $m$ books will be considered one thing. So, total $(n+1)$ things to arrange.

Number of ways

$=5:\left(1-\frac{1}{1 !}+\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}-\frac{1}{5 !}\right)$ $+4 !^3 C_1\left(1-\frac{1}{1 !}+\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}\right)$

$+3 !^3 C_2\left(1-\frac{1}{1 !}+\frac{1}{2 !}-\frac{1}{3 !}\right)+{ }^3 C_3 \cdot 1$

$=44+27+6+1=78$

Total number of digits used till $99=9+90 \times 2=189$

The digits use in next 610 three digit numbers $=1830$

$\therefore$ Total digit used till the number

$710=189+1830=2019$

Next three digits are $711$

$\therefore 2021$ st digit is $1 .$

abcde $\times 9=a b c d e(10-1)=a b c d e 0$

$-0 a b c d e$

$(a-1)-0=0 \Rightarrow a=1$

$10-e=a \Rightarrow e=9$

$b-a=e \Rightarrow[{b=0}]$

$(e-1)-d=b \Rightarrow[d=8]$

$d-c =c \Rightarrow [c=4]$

$\overline{a b c d e} =10989$

$\therefore$ Sum of digits $=27$

Out of given three letters $a, b, c$, we will repeat exactly one of the letter and we can do this in ${ }^3 C_1$ ways.

Now, we have four letters and out of these four letters two are identical, so number of ways to arrange these four letters is $\frac{4 !}{2 !}$.

So, the number of four letter words that can be formed with letters $a, b, c$ such that all three letters occur is

${ }^3 C_1 \times \frac{4 !}{2 !}=36$

The number of all four-letter words $=26^4$

The number of all four-letter words from set $V=\{a, e, i, o, u\}$ i.e vowels $=5^4$

Similarly, the number of all four-letter words from set of consonants $=21^5$

$\therefore$ Number of words which contains atleast one vowel and atleast one consonant

$=26^4-21^4-5^4$

$=\left(26^2+21^2\right)\left(26^2-21^2\right)-5^4$

$=(676+441)(5)(47)-5^4$

$=5[(1117 \times 47)-125]$

$=5 \times(52374)=261870$

Since in the group of first $10$ two digit number $10-19$, has atleast $1$ prime number similarly in other groups of 10 two digits numbers

$20-29$, 30-39, 40-49, 50-59, 60-69, 70-79, $80-89$ and $90-99$ have almost $1$ prime numbers.

So, the number of almost prime two-digit number is $90 .$

It is given that, the number of three digit number $\overline{a b c}$, such that

$\frac{b+c}{2}=b c$

the above relation is true if $b=c=0$

And if neither $b$ nor $c$ is zero, then

$\frac{1}{b}+\frac{1}{c}=2$, and $b, c \in\{1,2,3,4,5,6,7,8,9\}$

Then $b=c=1$

and $a \in\{1,2,3,4,5,6,7,8,9\}$

So, total number of such three digit number are $2 \times 9=18$

It is given that for prime numbers $p_1, p_2, p_3, p_4$ the special prime number

$p=p_1+p_2=p_3-p_4$

Case $I$

If all $p_1, p_2, p_3, p_4$ are odd, then $\left(p_1+p_2\right)$ and $\left(p_3-p_4\right)$ are even, which is not possible.

Case $II$

If one of $p_1$ and $p_2$ is even, say $p_2$ is $2$ and $p_4$ must be $2$ .

So, $p=p_1+2=p_3-2$

the above equation is satisfied only if

$p=5, p_1=3 \text { and } p_3=7$

So, the number of special prime $p$ is $1$ .

We have, $m$ is 5 -digits number using digits $1,2,3,4,5$ with repetition such that sum of two adjacent digit is odd and $n$ is 5 -digits number using digits $1,2$ and other is odd.

Even $=2,4$,Odd $=1,3,5$

Case $I$ Digit is repeated.

Two possibilities

$(a)$ odd even odd even odd

$=3 \times 2 \times 3 \times 2 \times 3=108$

$(b)$ even odd even odd even

$=2 \times 3 \times 2 \times 3 \times 2=72$

$\therefore \quad m=108+72=180$

Case $II$ Digit is not repeated.

The possibility of arrangement is odd even odd even odd

$=3 \times 2 \times 2 \times 1 \times 1=12$

$n=12$

$\frac{m}{n}=\frac{80}{12}=15$

Given,

$\frac{{ }^{n-1} C_5+{ }^{n-1} C_6 < { }^n C_7}{{ }^n C_6 < { }^n C_7}$

$\left[\because{ }^n C_{r-1}+{ }^n C_r={ }^{n+1} C_r\right]$

$\frac{n !}{(n-6) ! 6 !} < \frac{n !}{(n-7) ! 7 !}$

$n-6 > 7$

$n>13$

$\therefore$ Least value of $x=14$

Length of sides of trapezium from $\{1,2,3$, $4,5,6\}$.

From non-congruent trapezium

$|a-c| < b+d < a+c$

Possible combination are

$(r, p),(s, q) \equiv\{(5,6),(1,3)\},\{(5,6),(2,4)\}$

$\{(5,6),(1,4)\},\{(5,6),(3,4)\},\{(6,4),(1,3)\}$,

$\{(6,4),(1,5)\},\{(6,4),(2,3)\},\{(6,4),(3,5)\}$

$\{(4,5),(1,3)\}\{(4,5),(1,6)\}\{(4,5),(2,6)\}$

Total $11$ combination is possible.

Let triangle $T$ is $P Q R$ and other triangle is $A B C$.

$A$ can taken position if $\triangle A B C \sim \triangle P Q R$.

We can arrange $A, B, C$ in $3 !$ ways $=6$ ways

Total position of $A$ can take $=3 ! \times 2=12$ ways

Let $S$ be the set of all permutation in $a_1, a_2, a_3, a_4, a_5, a_6$ of $1,2 , 3,4,5,6$

$\therefore \quad S=6 !=720$

Let $S^{\prime}$ be the set of all permutation containing $(2,3,4)$.

Total number $=(2 !+3 !+4 !) 3$ !

$=(2+6+24) \times 6$

$=32 \times 6=192$

Number of element in $A=$ Number of clement in $S$-Number of element in $S^{\prime}$

$=720-192=528$

We have, $n \in\{2,3,4,5,6, \ldots, 200\}$ $\frac{1}{n}$ has terminating decimal of $n=2^a \times 5^b$ $\therefore n=2,4,5,8,10,16,20,25,32,40,50$, $64,80,100,125,128,160,200$

$\therefore$ Total number of $n=18$

We have,

$HCF (x, y)=16$

$\operatorname{LCM}(x, y)=48000$

We know,

Product of two number $= HCF \times LCM$

$\therefore \quad x y=16 \times 48000$

$x y=16 \times 16 \times\left(3^1 \times 2^3 \times 5^3\right)$

As $\operatorname{HCF}$ of $(x, y)=16$

$2^3$ can be selected in 1 ways and $3^1$ and $5^3$ can be selected in $(1+1)(3+1)=8$ ways

$\therefore$ Number of ordered pairs $=8$

Let two digits number

if $b$ is erased.

Then, the resulting number is $a$.

$\therefore a b$ is divisible by $a$ if $a b$ is multiple of $c$.

$\therefore$ Such number are $11,12,13,14,15,16$,

$17,18,19,22,24,26,28,33,36,39,44$,

$48,55,66,77,88,99$.

$\therefore$ Total number are $23$.

Hence, $\quad K < 25$

There are $90$ days from $1$ January to $31$ March (Non-leap year)

If year $13$ leap year, then total number of days $=91$ $(13$ weeks)

But we have $12$ Sunday

$\therefore 12$ weeks

$\therefore$ $1st$ Jan will be Monday as there will be 90 days January 1 to 31 March.

$\therefore 15$ th February will be Thursday.

Exponent of $2$ in $200 !$.

$=\left[\begin{array}{c}200 \\ 2\end{array}\right]+\left[\begin{array}{c}200 \\ 2^2\end{array}\right]+\left[\begin{array}{c}200 \\ 2^3\end{array}\right]+\left[\begin{array}{c}200 \\ 2^4\end{array}\right]+\left[\begin{array}{c}200 \\ 2^5\end{array}\right] +\left[\frac{200}{2^8}\right]+\left[\begin{array}{c}200 \\ 2^7\end{array}\right]+\left[\begin{array}{c}200 \\ 2^8\end{array}\right]$

$=100+50+25+12+6+3+1=197$

Exponent of $2$ in $100!$

$=\left[\frac{100}{2}\right]+\left[\frac{100}{2^2}\right]+\left[\begin{array}{c}100 \\ 2^3\end{array}\right]+\left[\begin{array}{c}100 \\ 2^4\end{array}\right]+\left[\begin{array}{c}100 \\ 2^5\end{array}\right]+\left[\begin{array}{c}100 \\ 2^6\end{array}\right]+\left[\begin{array}{c}100 \\ 2^7\end{array}\right]$

$\operatorname{In} \frac{200 !}{100 !}=\frac{2^{197}}{2^{97}}=2^{100}$

$\therefore$ The largest power of $2$ is $100$.

Let $p$ are positive number from $a_1, a_2, a_3, \ldots, a_n$

$\therefore n-p$ are negative number.

Given $a_j, a_k$ is positive $j < k$ and $a_j a_k=55$ $a_j a_k$ is positive.

$\therefore a_j$ and $a_k$ are both positive or negative.

$\therefore \quad{ }^P C_2+{ }^{n-P} C_2=55$

and $a_j a_k$ is negative $j < k$

$a_j a_k=50$

any one of $a_j$ and $a_k$ are positive:

$\therefore \quad{ }^P C_1 \times{ }^{n-P} C_1=50 \Rightarrow P(n-P)=50$

$\Rightarrow \quad{ }^P C_2+{ }^{n-P} C_2=55$

$\Rightarrow P(P-1)+(n-P)(n-P-1)=110$

$\Rightarrow P^2-P+(n-P)^2-(n-P)=110$

$\Rightarrow P^2+(n-P)^2-n=110$

$\Rightarrow\{P+(n-P)\}^2-2 P(n-P)-n=110$

$\Rightarrow n^2-100-n-110=0[\because P(n-P)=50]$

$\therefore \quad n^2-n-210=0$

$\Rightarrow \quad(n-15)(n+14)=0$

$n=15, n \neq-14$

$\therefore \quad P(15-p) =50$
$\Rightarrow p^2-15 p+50 =0 $

$(P-10)(P-5) =0$

$p=5 \text { or } 10$

$\therefore \quad p^2+(n-P)^2 =5^2+10^2$

$=25+100=125$

The distinct prime factor of

$43361=131 \times 331$

where $p_1=131$ and $p_2=331$

$p_1+p_2=131+331=462$

Given, $A=\{1,2,3, \ldots, 30\}$

Case $I$ All three number are multiple of $3$ then product of three number are divisible by $9 .$

$\therefore \quad{ }^{10} C_3=120$

Case $II$ Two number are multiple of $3$ and other are not multiple of $9$ .

i.e. ${ }^{10} C_2 \times{ }^{20} C_1=900$

Case $III$ One are multiple of $9$ and other two are not multiple of $3$ .

${ }^3 C_1 \times{ }^{20} C_2=570$

$\therefore$ Total number of ways $=120+900+570$ $=1590$

We have $2,4,6$ students from three schools respectively.

$6$ rooms allotted for $1,2,3,4,5,6$ such that each room has exactly two students for same schools.

$\therefore$ Total number of ways student can be accommodated in the rooms

We have,

Set $\{10,11,12, \ldots, 22\}$

Number of scalene triangle

$={ }^{13} C_3-3=283$

Number of isosceles triangle

$=\left({ }^{13} C_2 \times 2\right)-4=152$

Number of equilateral triangle

$={ }^{13} C_1=13$

So, total number of triangle

$=283+152+13=448$

Let two-digits number be

$n=10 a+b$

$\text { Given, } n =a^2+b^3$

$\therefore 10 a+b =a^2+b^3$

$\Rightarrow \quad a^2-10 a+b^3-b=0$

$\Rightarrow a(a-10)+b(b+1)(b-1)=0$

$\Rightarrow \quad b(b+1)(b-1)=a(10-a)$

$b \neq 1$ if $b=1$ then $a=10$ not possible

if $b=2, a(10-a)=6$ no value of $a$

$b=3, a(10-a)=24, a=4,6$

Numbers are $43$ and $63 .$

If $b=4, a(10-a)=60$ no value of $a$

If $b=5, a(10-a)=120$ not possible

$\therefore$ Numbers are $43$ and $63 .$