Maths M.C.Q (1 Marks)

Number of $2$ digit numbers ${ = ^4}{P_2}$

Number of $3$ digit numbers ${ = ^4}{P_3}$

Number of $4$ digit numbers ${ = ^4}{P_4}$

Hence the required number of ways

${ = ^4}{P_1}{ + ^4}{P_2}{ + ^4}{P_3}{ + ^4}{P_4}$.

$120(1 + 10 + {10^2} + {10^3}) = 120 \times 1111 = 133320$.

Number of $2$ digit numbers ${ = ^6}{P_2}$

Number of $3$ digit numbers ${ = ^6}{P_3}$

The required number of numbers $ = 6 + 30 + 120 = 156$.

start with $E$ are $720$, start with $MD$ are $5\;! = 120$ and

start with $ME$ are $120$.

Now the first word starts with $MO$ is nothing but $MODESTY$.

Hence rank of $= 720 + 720 + 120 + 120 + 1 = 1681$.

but the number of ways in which ${A_1}$ is above ${A_{10}}$ and the number of ways in which ${A_{10}}$ is above ${A_1}$ make up $10\;!$.

Also the number of ways in which ${A_1}$ is above ${A_{10}}$ is exactly same as the number of ways in which ${A_{10}}$ is above ${A_1}$.

Therefore the required number of ways $ = \frac{1}{2}(10\;!)$.

Now, remaining three places we have to fix up remaining three places we have to fix up remaining three which can be done in ${}^3{P_3}$ ways.

The total number of ways $ = {}^3{P_3} \times {}^3{P_3}= 36.$

Hence required no. of ways = $^3{P_3}$ $×$ $^4{P_4} = 144.$

Three vowels can be arrange at $6$ places in $^6{P_3} = 120$ ways. Hence the required number of arrangements $ = 120 \times 5\;! = 14400$.

$ = \frac{{30800}}{1} \Rightarrow 56 \times 55 \times (51 - r) = 30800$

$ \Rightarrow $ $r = 41$.

$\therefore $ The total number of arrangements

=$\frac{{7\,!}}{{2\,!\,2\,!\,1\,!\,1\,!\,1\,!}} = 1260$

But, the number of arrangements in which both $RR$ are together as one unit = $\frac{{6\,!}}{{2\,!\,1\,!\,1\,!\,1\,!\,1\,!}} = 360$

$\therefore $ The number of arrangements in which both $RR$ do not come together $= 1260 -360 = 900.$

{Since $IEEEENDPNDNC = 8$ letters}.

If we begin with a white ball, we find that $(n + 1)$ white balls numbered $1$ to $(n + 1)$ can be arranged in a row in $(n + 1)\;!$ ways.

Now $(n + 2)$ places are created between $n + 1$ white balls which can be filled by $(n + 1)$ black balls in $(n + 1)\;!$ ways.

So the total number of arrangements in which adjacent balls are of different colours and first ball is a white ball is $(n + 1)\;!\; \times (n + 1)\;!\; = {[(n + 1)\;!]^2}$.

But we can begin with a black ball also.

Hence the required number of arrangements is $2{[(n + 1)\;!]^2}$.

The total number of ways in which $2$ particular persons sit side by side is $10\;!\; \times \;2\;!$.

Hence the required number of arrangements $ = 11\;!\; - 10\;!\; \times 2\;!\; = 9 \times 10\;!$.

Similarly, each of $3$ and $5$ can be assigned in $3$ ways.

Thus, the no. of solutions are $3 × 3 × 3 = 27.$

Obviously required number of words are
$\frac{{6\;!}}{{2\;!}} \times 4\;!\; = 8640$.

Words starting from $I$ are $5 ! = 120$

Words starting from $KA$ are $4 ! = 24$

Words starting from $KI$ are $4 ! = 24$

Words starting from $KN$ are $4 ! = 24$

Words starting from $KRA$ are $3 ! = 6$

Words starting from $KRIA$ are $2 ! = 2$

Words starting from $KRIN$ are $2 ! = 2$

Words starting from $KRISA$ are $1 ! = 1$

Words starting from $KRISNA$ are $1 ! = 1$

Hence rank of the word $KRISNA$ is $324.$

(Note that a boy can stand first in all the four subjects).

Then two second prizes can be given in ways since a boy cannot get both the first and second prizes.

Hence the required number of ways $ = {20^4} \times {19^2}$.

We can arrange these letters in a line in $\frac{{5!}}{{2!}}$ ways.

In any such arrangement, $‘I’$ and $‘N’$ can be placed in $6$ available gaps in $^6{P_2}$ ways, so required number $ = \frac{{5!}}{{2!}}\,.{\,^6}{P_2}\, = \,{m_1}$.

Now, if word start with $‘ I ’$ and end with $‘R’$ then the remaining letters are $5$.

So total no. of ways $ = \frac{{5\,!}}{{2\,!}} = {m_2}$

$\therefore$  $\frac{{{m_1}}}{{{m_2}}} = \frac{{5!}}{{2!}}\,.\,\frac{{6!}}{{4!}}\,.\,\frac{{2!}}{{5!}}\, = \,30$.

therefore remaining members are $14$ but three special member constitute a member.

Therefore required number of arrangements are $12\;!\; \times 2$, because, chairman remains between the two specified persons and the person can sit in two ways.

Now no two female are to sit together and as such the $2$ female are to be arranged in five empty seats between two consecutive male and number of arrangement will be ${}^5{P_2}$.

Hence by fundamental theorem the total number of ways is

0= $4!\,\, \times \,\,{}^5{P_2}$   

$= 24 × 20 = 480$ ways.

$\therefore $$^\alpha {C_2}{ = ^{m(m - 1)/2}}{C_2} = \frac{1}{2}.\frac{{m(m - 1)}}{2}\left\{ {\frac{{m(m - 1)}}{2} - 1} \right\}$

$ = \frac{1}{8}m(m - 1)(m - 2)(m + 1)$

$ = \frac{1}{8}(m + 1)\;m(m - 1)(m - 2) = 3\;.{\;^{m + 1}}{C_4}$

Therefore the number of ways to fill up these places is ${2^{32}}$. As there is no person without a tooth, the maximum population is ${2^{32}} - 1$.

$\therefore $$\frac{7}{3}r - 1 = n - r = \frac{3}{2}(r + 1)$

or $14r - 6 = 9r + 9$ or $r = 3$. So, $n = 9$.

==>${}^{n + 1}{C_4} > {}^{n + 1}{C_3}\,\,(\,{}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r + 1}})$

==> $\frac{{{}^{n + 1}{C_4}}}{{{}^{n + 1}{C_3}}} > 1$

==> $\frac{{n - 2}}{4} > 1$

$ \Rightarrow n > 6$.

$ = 8 + 28 + 56 + 70 + 56 = 218$

{Since voter may vote to one, two, three, four or all candidates}.

$^n{C_1}{ + ^n}{C_2} + ......{ + ^n}{C_{n - 1}} = 254$

$ \Rightarrow $ ${2^n} - 2 = 254$

$ \Rightarrow $ ${2^n} = {2^8}$

.$ \Rightarrow $ $n = 8$.

${ = ^{12}}{C_1}{ + ^{12}}{C_2}{ + ^{12}}{C_3} + .......{ + ^{12}}{C_{12}} = {2^{12}} - 1$

$ = 4096 - 1 = 4095$.

$(10 + 1)(9 + 1)(7 + 1) - 1 = 11 \times 10 \times 8 - 1 = 879$.

$2$ bowlers and $9$ other players ${ = ^4}{C_2}{ \times ^9}{C_9}$

$3$ bowlers and $8$ other players ${ = ^4}{C_3}{ \times ^9}{C_8}$

$4$ bowlers and $7$ other players ${ = ^4}{C_4}{ \times ^9}{C_7}$

Hence required number of ways

$ = 6 \times 1 + 4 \times 9 + 1 \times 36 = 78$.

${ = ^4}{C_1}{ \times ^8}{C_5}{ + ^4}{C_2}{ \times ^8}{C_4}{ + ^4}{C_3}{ \times ^{8}}{C_3}{ + ^4}{C_4}{ \times ^8}{C_2}$

$ = 4 \times 56 + 6 \times 70 + 4 \times 56 + 1 \times 28 = 896$.

{Since $3$ vacancies filled from $5$ candidates in $^5{C_3}$ ways and now remaining

candidates are $22$ and remaining seats are $9$}.

$2$ girls and $5$ boys ${ = ^4}{C_2}{ \times ^6}{C_5} = 36$

$3$ girls and $4$ boys ${ = ^4}{C_3}{ \times ^6}{C_4} = 60$

Hence total ways $60 + 36 + 4 = 100$.

The two particular persons may be placed in two ways one in each boat. Therefore total number of ways are $ = 2{ \times ^8}{C_4}$.

$(i)$ Any one of the digits $1, 2, 3, 4$ repeats thrice and the remaining digits only once $i.e.$ of the type $1, 2, 3, 4, 4, 4$ etc.

$(ii)$ Any two of the digits $1, 2, 3, 4$ repeat twice and the remaining two only once $i.e.$ of the type $1, 2, 3, 3, 4, 4$ etc.

Now number of numbers of the $(i)$ type

$ = \frac{{6\;!}}{{3\;!}}{ \times ^4}{C_1} = 480$

Number of numbers of the $(ii)$ type

$ = \frac{{6\;!}}{{2\;!\;2\;!}}{ \times ^4}{C_2} = 1080$

Therefore the required number of numbers

$ = 480 + 1080 = 1560$.

The number of numbers from $1$ to $200$ which are not multiples of $5$ is $160$.

Therefore total number of two factor products which are not multiple of $5$ is $^{160}{C_2}$.

Hence the required number of factors ${ = ^{200}}{C_2}{ - ^{160}}{C_2} = 7180$.

But number of words on taking 5 different letters out of $10 = {\,^{10}}{C_5} = 252$

Required number of words = ${10^5} - 252 = 99748$.

$(i)$ Containing $3$ men and $3$ ladies

$(ii)$ Containing $2$ men and $4$ ladies

Required number of ways= ${(^8}{C_3}\, \times {\,^4}{C_3})\, + \,{(^8}{C_2}\, \times {\,^4}{C_4}) = 252$.

therefore in order to select one, two, three, …., $n$ coins.

Thus, if T is the total number of ways of selecting one coin, then

$T = {\,^{2n + 1}}{C_1}\,{ + ^{2n + 1}}{C_2}\, + \,...... + {\,^{2n + 1}}{C_n}\, = 255$…..$(i)$

Again the sum of binomial coefficients

${ = ^{2n + 1}}{C_0}{\mkern 1mu} { + ^{2n + 1}}{C_1}{\mkern 1mu}  + {{\mkern 1mu} ^{2n + 1}}{C_2} + ..... + $

$^{2n + 1}{C_n}{\mkern 1mu} { + ^{2n + 1}}{C_{n + 1}} + {{\mkern 1mu} ^{2n + 1}}{C_{n + 2}}{\mkern 1mu}  + ..... + $

$^{2n + 1}{C_{2n + 1}} = {(1 + 1)^{2n + 1}} = {2^{2n + 1}}$

$ =  = { > ^{2n + 1}}{C_0}{\mkern 1mu}  + {\mkern 1mu} 2{\rm{(}}2n + 1{C_1}{\mkern 1mu}  + {{\mkern 1mu} ^{2n + 1}}{C_2}{\mkern 1mu}  + ...{ + ^{2n + 1}}{C_n}{\rm{)}}$

$ + {{\mkern 1mu} ^{2n + 1}}{C_{2n + 1}}{\mkern 1mu}  = {2^{2n + 1}}$

==> $1 + 2(T) + 1 = {2^{2n + 1}} \Rightarrow 1 + T = \frac{{{2^{2n + 1}}}}{2} = {2^{2n}}$

==> $1 + 255 = {2^{2n}}\, \Rightarrow \,\,{2^{2n}}\,\, = \,\,{2^8}\, \Rightarrow \,\,n\,\, = \,\,4$.

Number of ways $ = \,{\,^5}{C_4} = 5$

Remaining Questions =$8$

Remaining Questions for solving =$6$

Number of ways ${\,^8}{C_6} = 28$

Number of total ways ${ = ^5}{C_4} \times {\,^8}{C_6}$ $ = 5 \times 28$ $ = 140$.

So if we put $n = 3$, then $r = 2$ satisfies the conditions.