Maths M.C.Q (1 Marks)

Given, $EDUCATION$ Vowel occurs in same order

$\text { _E_U_A_I_O- }$

There are $6$ place for letter $DCTN$ $\therefore$ Total number of arrangement is ${ }^6 C_4=15$

Let the number of red, white, blue, green balls be $r, w, b, g$ respectively and $r + w + b + g=n$.Given,

$\frac{{ }^r C_4}{{ }^n C_4}=\frac{{ }^w C_1 \times{ }^{\prime} C_4}{{ }^n C_4}$

$=\frac{{ }^w C_1 \times{ }^b C_1 \times{ }^r C_2}{{ }^n C_4}$
$=\frac{{ }^r C_1 \times{ }^{\infty} C_1 \times{ }^b C_1 \times{ }^5 C_1}{{ }^n C_4}$

$\Rightarrow \quad{ }^r C_4={ }^r C_3 \cdot{ }^w C_1$

$\Rightarrow \quad r-3=4 w \Rightarrow r=4 w+3$

${ }^r C_{ d }{ }^w C_1={ }^w C_1{ }^b C_1{ }^r C_2 \Rightarrow r=3 b+2$

${ }^r C_2{ }^w C_1^b C_1={ }^r C_1^w C_1^b C_1 C_1$

$\Rightarrow \quad r=2 g+1$

Now, LCM of $(4,3,2)=12$

$\therefore r_{\min }=11 \Rightarrow w_{\min }=2$

$\Rightarrow b_{\min }=3 \Rightarrow g_{\min }=5$

$\therefore$ Minimum number of ball

$=r + w + b + g=11+2+3+5=21$

$13 !=2 \times 3 \times 4 \times 5 \times 6 \times 7\times 8 \times 9 \times 10 \times 11 \times 12 \times 13$

$=2^{10} \times 3^5 \times 5^2 \times 7 \times 11 \times 13$

$24^k=\left(2^3 \times 3\right)^k$

When $13\,!$ is divide by $24^k$

$\therefore 2^{10} \times 3^5 \times 5^2 \times 7 \times 11 \times 13$

$\quad 2^{3 k} \cdot 3^k$

$\quad=2^{10-3 k} \cdot 3^{5-k} \cdot 5^2 \times 7 \times 11 \times 13$

$\therefore \quad 10-3 k=\text { integer }$

Then, maximum value of $k=3$

We have,

$S=\{(a, b): a, b \in Z, 0 \leq a, b \leq 18\}$

Number of line in $R^2$ passing through $(0,0)$ and one other point in $S$ is $34 .$

$(1,17),(2,17),(3,17), \ldots,(18,17)=17$ points

$(17,1),(17,2),(17,3), \ldots,(17,18)=17$ points

$\therefore$ Total number of lines $=34$

We have,

$(A, B)$ is subset of $X$, where $X$ has 5 elements $A \neq \phi, B \neq \phi, A \cap B=\phi$

The order pairs if $(A, B)$ is

${ }^5 C_2 \times 2 !+{ }^5 C_3\left({ }^3 C_2 \times 2 !\right)+{ }^5 C_4\left({ }^4 C_3 \times 2\right.$

$\left.+{ }^4 C_2\right)+{ }^5 C_5\left({ }^5 C_4 \times 2 !+{ }^5 C_3 \times 2 !\right)$

$=20+60+70+30=180$

$\therefore 151 \leq d \leq 200$

Hence, option $(c)$ is correct.

We have,

$A_n =\max \left\{{ }^n C_r \mid 0 \leq r \leq n\right\}$

$n \in\{1,2,3, \ldots, 20\}$

Case $I$ When $n$ is even

$A_n={ }^n C_{n / 2}$

$\therefore \quad \frac{A_n}{A_{n-1}}=\frac{{ }^n C_{n ' 2}}{{ }^{n-1} C_{\frac{n-1-1}{}}^2}=2$

So for all $n$ even given relation is true.

Case II When $n$ is odd

$A_n={ }^n C_{\frac{n-1}{}}^2$

$\therefore \quad \frac{A_n}{A_{n-1}} =\frac{{ }^n C_{n-1}}{{ }^{n-1} C_{n-1}}=\frac{2 n}{n+1}$

$19 \leq \frac{2 n}{n+1} \leq n \text { if } n=19$

$\therefore$ Total number of elements are 10 even number and $19=10+1=11$

We have,

$S=\{(a, b): a, b, \in Z, 0 \leq a, b \leq 18\}$

$3 x+4 y+5(x, y) \in S$

$S \leq 3 x+4 y+5 \leq 131$

$[\because \min (x, y)=(0,0), \max (x, y)=(15,18)]$

Given, $3 x+4 y+5$ is divisible by 19

$\therefore 3 x+4 y+5=19,38,57,76,95,114$

Case $I$

$\begin{array}{r}3 x+4 y+5=19 \\3 x+4 y=14\end{array}$

Only $(2,2)$ satisfies

Case $II$

$\begin{array}{r}3 x+4 y+5=38 \\3 x+4 y=33\end{array}$

Possible values of $(x, y)$ is $(3,6),(7,3)$, $(11,0)$

Case $III$

$\begin{array}{r}3 x+4 y+5=57 \\3 x+4 y=52\end{array}$

Possible values of $(x, y)$ is $(0,13),(4,10)$, $(8,7),(12,4),(16,1)$

Case $IV$

$3 x+4 y+5=76$

Possible values of $(x, y)$ is $(1,17),(5,14)$,

$(9,11),(13,8),(17,5)$

Case $V$

$3 x+4 y+5=95$

Possible values of $(x, y)$ is

$(6,18),(10,15),(14,12),(18,9)$

Case $VI$

$\begin{aligned}3 x+4 y+5 &=114 \$x, y) &=(15,15)\end{aligned}$

$\therefore$ Total solution $=19$

We have,

$6-$digits number are $a b a b a b$.

$\because a b a b a b=10^5 a+10^4 b+10^3 a+10^2 b+10a+b$

$=\left(10^5+10^3+10\right) a+\left(10^4+10^2+1\right) b$

$=\left(10^4+10^2+1\right)(10 a+b)$

$=(10000+100+1)(10 a+b)$

$=(10101)(10 a+b)$

$=3 \times 7 \times 13 \times 37(10 a+b)$

Since, $6-$digit number are product of exactly $6$ primes.

$\because 10 a+b$ is product of $2$ primes,

$10 a+b$ is lie between $10$ to $99$

$10 a+b=10=2 \times 5$

$22=2 \times 11$

$34=2 \times 17$

$38=2 \times 19$

$46=2 \times 23$

$55=5 \times 11$

$58=2 \times 29$

$62=2 \times 31$

$74=2 \times 37$

$82=2 \times 41$

$85=5 \times 17$

$94=2 \times 47$

$95=5 \times 19$

$\because 13,6$-digits number.

We have, $a_2, a_3, a_4, a_5, a_6, a_7$ are integers.

$\frac{5}{7}=\frac{a_2}{2 !}+\frac{a_3}{3 !}+\frac{a_4}{4 !}+\frac{a_5}{5 !}+\frac{a_6}{6 !}+\frac{a_7}{7 !}$

and $0 \leq a_j < j$

$5=\frac{2520 a_2+840 a_3+210 a_4+42 a_5+7 a_6+a_7}{7 !}$

$\Rightarrow 3600=2520 a_2+840 a_3+210 a_4 +42 a_5+7 a_6+a_7$

$0 \leq a_j < j a_2 =1 a_3 \in\{1,2\}$

If $a_3=2$, then $2520+(840) \times 2 > 3600$

$\because a_3$ must be $1$

$a_4 \in\{1,2,3\}$

If $a_4=2$, then $2520+840+210(2) > 3600$ $\because a_4$ must be $1$

$\because \quad 3600=2520+840+210+42 a_5+7a_6+a_7$

$30=42 a_5+7 a_6+a_7$

$a_5 \in\{1,2,3,4\}$

$if\,\,a_5=1$

$30 < 42+7 a_6+a_7$

$a_5=0$

Put $a_5=0$, then $30=7 a_6+a_7$ $\because a_6=4$ and $a_7=2$

$\because a_2+a_3+a_4+a_5+a_6+a_7=1+1+1+0+4+2=9$

abcis three-digits number $a b c=100 a+10 b+c$

$100 \leq a b c<999, a \in\{1,2,3, \ldots, 9\} b, c \in\{0$, $1,2,3, \ldots, 9\}$

Now, $(a \times b \times c)+(a \times b)+(b \times c) +(c \times a)+a+b+c=29$

$(a \times b)(c+1)+b(c+1)+a(c+1) +(c+1)=30$

$\Rightarrow(c+1)(b(a+1)+1(a+1))=30$

$\Rightarrow \quad(a+1))(b+1)(c+1)=30$

$\leq a \leq 9,0 \leq b, c \leq 9$

$\because$ Total number of solution $=18$

We have,

A polygon of $10$ sides.

$\therefore$ Sum of total interior angle of $10$ sided polygon $=(10-2) \times \pi=8 \pi$

Let $k$ be the number of interior angle of $P$ that are greater than $180^{\circ}$.

$\therefore \quad h \times 180^{\circ}$

$\therefore \quad k \pi+x=8 \pi$

$\therefore$ Maximum value of $k$ is $7, x$ is the sum of three angles can be less than $\pi$.

We have $10$ points lie a plane such that no three of them are collinear.

According to question only $5$ ways are

possible i.e. $1-6,2-7,3-8,4-9$ and $5-10$

We have,

Five teams in a tournament and each team plays against every team.

$\therefore$ Total number of one team $=4$

Possible number of distribution

We have,

Integer $n$ with $100 \leq n \leq 999$

$n=999-99=900$

$n$ containing three distinct digits are $9 \times 9 \times 8=648$

$n$ containing at most two distinct digit is

Total number $-n$ containing three

distinct digits $=900-648=252$

We have,

Three children with their guardians

$\therefore$ Total number of persons $=6$

Total number of ways when principal want to interview all the 6 persons such that no child interviewed before their guardians is $\frac{6 !}{2 ! 2 ! 2 !}=\frac{720}{8}=90$ ways

We have,

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

$m n =144$

$m n =2^4 \times 3^2$

Total number of divisor of $144$ is $(4+1)(2+1)=15$

When $m$ and $n$ are positive integers.

If $m$ and $n$ are negative integers, then also number of divisor is $15 .$

$\therefore$ Total ordered pairs of $(m, n)$ when $m$ and $n$ are integers $=15+15=30$

Let the student answered correct $=x$

Student answer wrong $=y$

Student unattempted $=z$

According to the question,

$x+y+z=30$, and $4 x+z=60$

$x=15, y=15, z=0$

$x=14, y=12, z=4$

$x=13, y=9, z=8$

$x=12, y=6, z=12$

$x=11, y=3, z=16$

$x=10, y=0, z=20$

Total number of cases $=6$

Digit $1$ appears in $1,10,11$, and $12$ in hour.

$\therefore$ The clock will show the incorrect time between $1-2,10-11,11-12,12-1$ day and night both incorrect time $(8 \times 60)=480\,min$

Digit $1$ appear in minutes $1,10,11,12$,

$13,14,15,16,17,18,19,21,31,41,51$

$=15\,min$

$\therefore$ It will shows the incorrect time

$=16 \times 15$

$=240\,min$

Total incorrect time $=240+480$

$=720\,min$

Correct time $=24 \times 60-720$

Fraction of correction time

$\frac{=24 \times 60-720}{24 \times 60}$

$=\frac{1}{2}$

$6-$digit number are there in which no digits is repeated and even digits appear on even places and odd digits appear in odd place. Such $6$ digit number which is divisible by $4$. If last two digits are divisible by $4$

i.e. $12,16,32,36,52,56,72,76,92,96$

$\therefore$ Last digits appears only $2$ and $6$

$3$ way $3$ ways $4$ ways $4$ way $5$ way $2 / 6$

$Total numbers =3 \times 3 \times 3,5,7,9$

$=1440$

We have,

$12$ -sided regular polygon the diagonal of rectangle are equal.

$\therefore$ Diagonals of rectangle passes through the centre of $12$ -sided regular polygon $=6$

$\therefore$ Number of rectangles $={ }^6 C_2=\frac{6 !}{2 ! 4 !}=15$

Total number of game $=9$

Total number of player $=3$

Total score be zero

If every team wins $3$ games and lose their $6$ games, then final score be zero.

$\therefore$ Total number of ways

${ }^9 C_3 \times{ }^6 C_2 \times{ }^3 C_3$

$\Rightarrow \quad \frac{9 !}{3 ! 6 !} \times \frac{6 !}{3 ! 3 !} \times 1=\frac{9 !}{3 ! 3 ! 3 !}=1680$

An envelope has space for at most 3 stamps. Three stamps of denomination is 1 and three stamps of denomination is $a$.

The value of stamps of envelope must be

$1+1+1=3, a+a+a=3 a$

$a+a+1=2 a+1, a+1+1=a+2$

$a > 1, \text { If } a=2$

then maximum value of stamps is

$3 \times 2=6$

$\therefore$ Least positive integer has not stamp value is $7 .$

We have, $\frac{(101)^{k / 2}}{k !}$

$\sqrt{10} \overline{1} > 10$

$\frac{(\sqrt{101})^9}{9 !}-\frac{(\sqrt{10}) 1)^{10}}{10 !}$

$\frac{(\sqrt{10} 1)^9}{9 !}\left[1-\frac{\sqrt{10} \overline{0}}{10}\right] < 0 \quad\left[\because \frac{\sqrt{101}}{10} > 1\right]$

$\therefore \quad \frac{(\sqrt{10} \overline{1})^9}{9 !} < \frac{(\sqrt{101})^{10}}{10 !}$

$\frac{(\sqrt{101})^{10}}{10 !}-\frac{ \quad(\sqrt{101})^{11}}{11!}$

$=\frac{(\sqrt{10} \overline{0})^{10}}{10 !}\left[1-\frac{\sqrt{10} \overline{1}}{11} > 0\left[\because \frac{\sqrt{101}}{11} < 1\right]\right.$

$\frac{(\sqrt{101})^{10}}{10 !} > \frac{(\sqrt{101})^{11}}{11 !}$

$=\frac{\sqrt{101}}{10 !}$ $\left[1-\frac{(\sqrt{101})^{91}}{11 \times 12 \ldots 101}\right] > 0$ 

$\therefore$ For $k=10$, Maximum value of $\frac{(101)^{k / 2}}{k !}$

$5,0,0,0 \Rightarrow 1$ Ways

$4,1,0,0 \Rightarrow \frac{5 !}{4 !}=5$ 

$3,2,0,0, \Rightarrow \frac{5 !}{3 ! 2 !}=10$ ways

$2,2,0,1 \Rightarrow \frac{5 !}{2 ! 2 ! 2 !}=15$ ways

$2,1,1,1 \Rightarrow \frac{5 !}{2 !(1 !)^3 3 !}=10$ ways

$3,1,1,0 \Rightarrow \frac{5 !}{3 ! 2 !}=10$ ways

Total $\Rightarrow 1+5+10+15+10+10=51$ ways

$ \mathrm{B} \_4 !=24 $

$ \mathrm{H} \_\frac{4 !}{2 !}=12 $

$ \mathrm{~J} \_\frac{4 !}{2 !}=12$

$O B B H J$

$O B B J H$ $\rightarrow 50^{\text {th }}$ rank

$A \rightarrow 5 !=120$

$\mathrm{G} \mathbb{\circledR} 5 !=120 \quad 240$

NA $\mathbb{\circledR} 4 !=24 \quad 264$

$\mathrm{NG} \mathbb{\circledR} 4 !=24 \quad 288$

$N P {\circledR} 4 !=24 \quad 312$

$NRAGPU$ $=1 \quad 313$

$NRAGUP $$\quad 314$

$NRAPGU $ $\quad 315$

total number of three digit numbers not divisible by $3$ will be formed by using the digits

$ (4,5,7) $

$ (3,4,7) $

$ (2,5,7) $

$ (2,4,7) $

$ (2,4,5) $

$ (2,3,5) $

number of ways $=6 \times 3 !=36$

$=\frac{17 \times 16}{2}=136$

$\mathrm{L}_2, \mathrm{~L}_4, \mathrm{~L}_6,--\mathrm{L}_{20}$ are Concurrent

Total points of intersection $={ }^{20} \mathrm{C}_2-{ }^{10} \mathrm{C}_2-{ }^{10} \mathrm{C}_2+1$

$=101$

$ =751$

$=\frac{{ }^8 \mathrm{C}_1 \cdot{ }^4 \mathrm{C}_2}{3}=16$

$ \alpha=\frac{(24) !}{(4 !)^6}, \beta=\frac{(120) !}{(5 !)^{24}}$

Let $24$ distinct objects are divided into $6$ groups of $4$ objects in each group.

No. of ways of formation of group $=\frac{24 !}{(4 !)^6 \cdot 6 !} \in \mathrm{N}$

Similarly,

Let $120$ distinct objects are divided into $24$ groups of $5$ objects in each group.

No. of ways of formation of groups

$=\frac{(120) !}{(5 !)^{24} \cdot 24 !} \in N$

Words starting with $\mathrm{GE}=60$

Words starting with $\mathrm{GN}=60$

Words starting with $GTE$ $=24$

Words starting with $GTN$ $=24$

Words starting with $GTT$ $=24$

GTWENTY $=1$

Total $=553$

$2$ shelf empty : $\left.\begin{array}{c}(7,1,0,0) \\ (6,2,0,0) \\ (5,3,0,0) \\ (4,4,0,0)\end{array}\right] \rightarrow 4$ ways

$1$ shelf empty: $\left.: \begin{array}{cc}(6,1,1,0) & (3,3,2,0) \\ (5,2,1,0) & (4,2,2,0) \\ (4,3,1,0) & \end{array}\right] \rightarrow 5$ ways

$0$ Shelf empty : $\left.\begin{array}{ll}(1,2,3,2) & (5,1,1,1) \\ (2,2,2,2) \\ (3,3,1,1) & \\ (4,2,1,1) & \end{array}\right] \rightarrow 5$ ways

total = $15$ ways

Number of words with selection ( $a, a, a, b)$

$={ }^8 \mathrm{C}_1 \times \frac{4 !}{3 !}=32$

Number of words with selection $(a, a, b, b)$

$=\frac{4 !}{2 ! 2 !}=6$

Number of words with selection ( $a, a, b, c)$

$={ }^2 C_1 \times{ }^8 C_2 \times \frac{4 !}{2 !}=672$

Number of words with selection $(a, b, c, d)$

$={ }^9 \mathrm{C}_4 \times 4 !=3024$

$\therefore \text { total }=3024+672+6+32$

 $=3734$

($i$)  All distinct digits

$ a+b+c=14 $

$ a \geq 1 $

$ b, c \in\{0 \text { to } 9\}$

by hit and trial : $8$ cases 

$(6,5,3)$ $(8,6,0)$ $(9,4,1)$ 

$(7,6,1)$ $(8,5,1)$ $(9,3,2)$ 

$(7,5,2)$ $(8,4,2)$ 

$(7,4,3)$ $(9,5,0)$

($ii$)  $2$ same, $1$ diff $\quad a=b$;

$2 \mathrm{a}+\mathrm{c}=14$

by values :

$\left.\begin{array}{l}(3,8) \\ (4,6) \\ (5,4) \\ (6,2) \\ (7,0)\end{array}\right\} \frac{3 !}{2 !} \times 5-1$

$=14$ cases

 $(iii)$ all same

$ 3 a=14 $

$ a=\frac{14}{3} \times$ rejected

$0$ cases

Hence, Total cases :

$ 8 \times 3 !+2 \times(4)+14 $

$ =48+22 $

$ =70$

$\underbrace{r+1 \geq 0, \quad r \geq 0}_{r \geq 0}$

$\frac{{ }^{n-1} C_{\mathrm{r}}}{{ }^n C_{\mathrm{r}+1}}=\mathrm{k}^2-8$

$\frac{\mathrm{r}+1}{\mathrm{n}}=\mathrm{k}^2-8$

$\Rightarrow \mathrm{k}^2-8>0$

$(\mathrm{k}-2 \sqrt{2})(\mathrm{k}+2 \sqrt{2})>0$

$\mathrm{k} \in(-\infty,-2 \sqrt{2}) \cup(2 \sqrt{2}, \infty)$   $......(I)$

$\therefore \mathrm{n} \geq \mathrm{r}+1, \frac{\mathrm{r}+1}{\mathrm{n}} \leq 1$

$ \Rightarrow \mathrm{k}^2-8 \leq 1 $

$\mathrm{k}^2-9 \leq 0$

$ -3 \leq \mathrm{k} \leq 3 $                              $......(II)$

From equation $(I)$ and $(II)$ we get

$\mathrm{k} \in[-3,-2 \sqrt{2}) \cup(2 \sqrt{2}, 3]$

Remaining $3$ questions can be selected either in $(1,1,1)$ or $(3,0,0)$ or $(2,1,0)$

$\therefore$ Total ways $={ }^8 \mathrm{c}_5 \cdot{ }^6 \mathrm{c}_5 \cdot{ }^6 \mathrm{c}_5+{ }^8 \mathrm{c}_6{ }^6 \mathrm{c}_5 \cdot{ }^6 \mathrm{c}_4 \times 2+$

${ }^8 \mathrm{c}_5 \cdot{ }^6 \mathrm{c}_6 \cdot{ }^6 \mathrm{c}_4 \times 2+{ }^8 \mathrm{c}_4 \cdot{ }^6 \mathrm{c}_6 \cdot{ }^6 \mathrm{c}_5 \times 2+{ }^8 \mathrm{c}_7 \cdot{ }^6 \mathrm{c}_4 \cdot{ }^6 \mathrm{c}_4 $

$ =56 \cdot 6 \cdot 6+28 \cdot 6 \cdot 15 \cdot 2+56 \cdot 15 \cdot 2+70 \cdot 6 \cdot 2 $

$ +8 \cdot 15 \cdot 15 $

$ =2016+5040+1680+840+1800=11376$

${ }^7 \mathrm{C}_{\mathrm{m}+1}+{ }^7 \mathrm{C}_{\mathrm{m}+2}>{ }^8 \mathrm{C}_3$

${ }^8 \mathrm{C}_{\mathrm{m}+2}>{ }^8 \mathrm{C}_3$

$\therefore \mathrm{m}=2$

$\text { And }{ }^{\mathrm{n}-1} \mathrm{P}_3:{ }^{\mathrm{n} P_4=1: 8}$

$\frac{(n-1)(n-2)(n-3)}{n(n-1)(n-2)(n-3)}=\frac{1}{8}$

$\therefore \mathrm{n}=8$

$\therefore{ }^{\mathrm{n}} \mathrm{P}_{\mathrm{m}+1}+{ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{m}}={ }^8 \mathrm{P}_3+{ }^9 \mathrm{C}_2$

$=8 \times 7 \times 6+\frac{9 \times 8}{2}$

$=372$

($1$) All distinct

${ }^8 \mathrm{C}_5 \rightarrow 56$

($2$) $2$ same, $3$ different

${ }^3 \mathrm{C}_1 \times{ }^7 \mathrm{C}_3 \rightarrow 105$

($3$) 2 same $I^{\text {st }}$ kind, 2 same $2^{\text {nd }}$ kind, 1 different

${ }^3 \mathrm{C}_2 \times{ }^6 \mathrm{C}_1 \rightarrow 18$

Total $\rightarrow 179$

$\frac{4 !}{2 ! 2 !} \times \frac{5 !}{2 ! 3 !}=\frac{24 \times 120}{4 \times 12}=60$

$=2 \times 4 \times 3 \times 2=48$

Five digit number $=5 !=120$

Total number greater than $7000$

$=120+48=168$

Using digits $1,3,5,7,9$

Total Numbers $=3 \times 4 \times 3 \times 2=72$

$ C \ldots \ldots \ldots \ldots \ldots \ldots \rightarrow 5 !=120 $

$ I \ldots \ldots \ldots \ldots \ldots \ldots \rightarrow 5 !=120$

$L \ldots \ldots \ldots \ldots \ldots \ldots 5 !=120 $

$ PB \ldots \ldots \ldots \ldots \ldots \rightarrow 4 !=24 $

$ PC \ldots \ldots \ldots \ldots \ldots \ldots \rightarrow 4 !=24 $

$ PL \ldots \ldots \ldots \ldots \ldots \rightarrow 4 !=24 $

$ PI \ldots \ldots \ldots \ldots \ldots \rightarrow 4 !=24$

$ P \cup BC \ldots \ldots \ldots \rightarrow 2 !=2$

$ P \cup BI \ldots \ldots \ldots \rightarrow 2 !=2$

$ P \cup BLC \ldots \ldots . \rightarrow 1 !=1 $

$ P \cup BLIC \ldots \ldots \rightarrow=1 $

 Serial number $=4(120)+4(24)+6=582$

$NNN , DD , P , C$

$\frac{8 !}{3 ! 2 !} \times \frac{6 !}{41}=16800$

$\Rightarrow n=8$

Therefore total persons $=16$

$=\frac{8 \times 7 \times 6 \times 5 \times 4}{4}$

$=56 \times 30$

$=1680$