Permutation and Combination Exercise 3.6

                         Exercise 3.6

1.Find the value of

$a)^{15}C_4$

Answer-Here n=15, r=4                     

as,$^nC_r=\frac{n}{r!(n-r)!}$

$=\frac{15!}{4!(15-4)!}$

$=\frac{15.14.13.12.11!}{4.3.2.1}$

$=5.7.13.3$

=1365

b)$^{80}C_2$

Answer- n=80 ,n=2  

    as, $^nC_r=\frac{n}{r!(n-r)!}$

=$\frac{80!}{2!(80-2)!}$

$=\frac{80!}{2!78!}$

$=\frac{80\times79\times78!}{2\times78!}$

=40.79

=3260

$c)^{15}C_4+^{15}C_4$

Answer-

As, $^nC_r+^nC_r=^{n+1}C_r$

So, n=15, r=4

$\therefore ^{15}C_4+^{15}C_4$

$=^{16}C_5$

d)$^{20}C_{16}+^{19}C_{16}$

Answer-

$=^{19}_C16+^{19}C_15-{19}C_16$

$=^{19!}C_{15}$

$=\frac{19!}{(19-15)!}$

$=\frac{19!}{15!4!}$

$=\frac{19.18.17.16.15!}{15!.4.3.2.1}$

=19.6.17.2

=3876

2.Find n if

a) $^6P_2=n ^6C_2$

Answer-

$\frac{6!}{(6-2)!}=n\frac{6!}{2!4!}$

$\frac{6\times5\times4!}{4!}=n\frac{6\times5\times4!}{2!\times4!}$

30=n(3.5)

So, n=2

b)$^{2n}C_3:^nC_2=\frac{52}{3}$

Answer-

$\therefore\frac{^{2n}C_3}{^nC_2}=\frac{52}{3}$

$\therefore\frac{\frac{2n!}{3!(2n-3)!}}{\frac{n!}{3!(n-2)!}}=\frac{52}{3}$

$\therefore\frac{2n!}{3!(2n-3)!}\times\frac{2!(n-2)!}{n!}=\frac{52}{3}$

$\therefore\frac{2n(2n-1)(2n-2)(2n-3)!2(n-2)!}{3!(2n-3)!n(n-1)(n-2)!}=\frac{52}{3}$

$\therefore 4(2n-1)=52$

$\therefore 2n-1=13$

$\therefore n=7$

c) $^{n}C_{n-3}=84$

Answer-

$\frac{n!}{(n-3)!3!}=84$

$\frac{n(n-1)(n-2)(n-3)!}{(n-3)!3!}=84$

$\frac{n(n-1)(n-2)}{3\times2\times1}=84$

$n(n-1)(n-2)=8\times9\times7$

$n=9$

 3.Find r if

a)$^{14}C_{2r}:^{10}C_{2r-4}=\frac{143}{10}$

Answer-

$\frac{\frac{14!}{(2r)!(4-2r)!}}{\frac{10!}{((2r-4)!(14-2r)!}}$=$\frac{143}{10}$

$\frac{14!}{(2r)!(4-2r)}!$$\times$$\frac{((2r-4)!(14-2r)!}{10!}$

$\frac{14.13.12.11.10!(2r-4)!}{2r(2r-1)(2r-2)(2r-3)(2r-4)!10!}=143:10$

$\frac{14.13.12.11}{2r(2r-1)(2r-2)(2r-3)}=143:10$

$r(2r-1)(2r-2)(2r-3)=\frac{14.13.12.11.10}{143.2}$

$r(2r-1)(2r-2)(2r-3)=720$

$r(2r-1)(2r-2)(2r-3)=10.9.8$

$r=10$

4. Find n and r if 

a) $^{n}P_{r}$ and $^{n}C_{n-r}=120$

Answer-

$^nP_r$    $^nC_n-r=120$

$\frac{n!}{(n-r)!}=720  ....(i)       \frac{n!}{(n-r)!r!}=120$  ....(ii)

dividing by (i) and (ii)

$\frac{n!}{(n-r)!}\times\frac{r!(n-r)!}{n!}=\frac{720}{120}$

                r!=6

so r=3

b)$^nC_{r-1}:^nC_r:^nC_{r+1}=20:35:42$

Answer-

 As, $^nC_{r-1}:^nC_r=\frac{20}{35}$

 $\frac{\frac{n!}{(r-1)!(n-r+1)!}}{\frac{n!}{r!(n-r)!}}$=$\frac{20}{35}$ 

 $\frac{n!}{(r-1)!(n-r+1)!}\times \frac{r!(n-r)!}{n!}$=$\frac{20}{35}$

 $\frac{r(r-1)!(n-r)!}{(r-1)!(n-r+1)!(n-r)!}$=$\frac{20}{35}$

 $\frac{r}{n-r+1}=\frac{20}{35}$

 35r=20(n-r+1)

 11r-4n=4  .....(i)

 now, $^nC_r:^nC_{r+1}=35:42$

 $\frac{\frac{n!}{r!(n-r)!}}{\frac{n!}{(r+1)!(n-(r+1))!}}$=$\frac{35}{42}$

 $\frac{n!}{r!(n-r)!} \times \frac{(r+1)!(n-(r+1))!}{n!}$=$\frac{35}{42}$

 $\frac{(r+1)r!(n-r-1)!}{r!(n-1)(n-r-1)!}$=$\frac{35}{42}$

 $\frac{r+1}{n-r}=\frac{35}{42}$

 $\therefore$ 42(r+1)=35(n-r)

 $\therefore$ 42r+42=35n-35r

 11r-5n=-6 ....(ii)

 So,    11r-4n=4

          11r-5n=-6

solving above equation we get the value of n=10

 and now substituting the value of n= 10 in eq (i) we get,

 11r-5(10)=-6

 11r=44

  r=4 and n=10

5. If  $^nP_r=1814400$ and $^nC_r=45$ find $^{n+4}C_{r+3}$

Answer-

$\frac{^nP_r}{^nC_r}$=$\frac{1814400}{45}$

$\frac{\frac{n!}{(n-r)!}}{\frac{n!}{r!(n-r)!}}$=$\frac{1814400}{45}$

$\frac{n!}{(n-r)!}\times \frac{r!(n-r)!}{n!}$=$\frac{1814400}{45}$

$\frac{n!r!(n-r)!}{(n-r)!n!}$=$\frac{1814400}{45}$

r!=40320

r!=8.7.6.5.4.3.2.1

so, r=8

Also, $^nC_r=45$

$^nC_8=45$

$=\frac{n!}{8!(n-8)!}$

$n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)(n-7)=10.9.8.7.6.5.4.3.2.1$

Compairing on both sides, we get

n=10

$^{n+4}C_{r+1}=^{14}C_{11}$

$=\frac{14!}{11!(14-11)!}$

$^{14}C_{11}=364$ 

6. If  $^nC_{r-1} =6435$, $^nC_r=5005$, $^nC{r+1}=3003$ Find $^{r}C_3$

Answer-

$\frac{^nC_{r-1}}{^nC_r}=\frac{6435}{5005}$

$\frac{\frac{n!}{(r-1)!(n-r+1)!}}{\frac{n!}{r!(n-r)!}}$=$\frac{6435}{5005}$

$\frac{n!}{(r-1)!(n-r+1)!} \times \frac{r!(n-r)!}{n!}=$\frac{6435}{5005}$

$\frac{r(r-1)!(n-r)!}{(r-1)!(n-r+1)(n-r)!}$=$\frac{6435}{5005}$

$\frac{r}{(n-r+1)}$==$\frac{6435}{5005}$

$\frac{r}{n-r+1}$=$\frac{9}{7}$

7r=9n-9r+9

16r-9n=9            ....(i)

$\frac{^nC_r}{^nC_{r+1}}=\frac{5005}{3003}$

$\frac{\frac{n!}{r!(r-1)!}}{\frac{n!}{(r+1)!(n-r-1)!}}$=$\frac{5}{3}$

$\frac{n!}{r!(r-1)!}\frac{(r+1)!(n-r-1)!}{n!}$=$\frac{5}{3}$

$\frac{(r+1)r!(n-r+1)!}{r!(n-r)!n!}$=$\frac{5}{3}$

$\frac{r+1}{n-r}$=$\frac{5}{3}$

3(r+1)=5(n-r)

8r-5n=-3            ....(ii)

now, solving these equation

 

16r-9n=9

8r-5n=-3

r=9

$^rC_5=^9C_5$

$=\frac{9!}{5!4!}$

=126 

7.Find the number of ways of drawing 9 balls from a bag that has 6 red balls 8 green balls and 7 blue was so that three balls of every colour are drawn.

Answer-

Total number ways of balls=8+5+7=18

Number of red balls=6

Number of green balls=5

Number of blue balls=7

9 balls are to be drawn 3 of each colour,

No. of ways of drawing 3 red balls of 6 red balls=$^6C_3$

similarly 3 green balls out of 5 geen balls=$^5C_3$

3 blue ball out of 7 blue balls=$^7C_3$

\therefore $total number ways$=$^6C_3\times^5C_3\times^7C_3$

=$\frac{6!}{3!3!}\times \frac{5!}{3!2!} \times \frac{7!}{3!4!}$

$=20 \times 10 \times 35$

=7000 

8.It. find the number of ways of selecting a team of 3 boys and 2 girls from 6 boys and 4 girls. 

Answer-

there are 6 boys and 4 girls

we have to select 3 boys out of 6 boys=$^6C_3$

we have to select 2 girls out of 4 girls=$^4C_2$

$\therefore$ total number of ways

=$^6C_3 \times ^4C_2$

$=\frac{6!}{3!3!} \times \frac{4!}{2!2!}$

$=20\times 6$ 

9. After meeting every participant shake hands with every other participant. If the number of handshakes is 66 find the number of participants in the meeting.

Answer- 

 Let there be n participants present in the meeting.                     

 A handshake occurs between 2 persons.      

 ∴ Number of handshakes = nC2.       

 Given 66 handshakes were exchanged.          

 ∴ 66 = nC2                                                     

 ∴ 66 =$\frac{n!}{2!(n-2)!}$               

 ∴ 66 × 2 =$\frac{n(n-1)(n-2)!}{(n-2)!}$

 ∴ 132 = n(n – 1)                                                

 ∴ n(n – 1) = 12 × 11                             

 Comparing on both sides, we get                     

 n = 12                                                                     

 ∴ 12 participants were present at the meeting.

10. If 20 points are marked on a circle how many chords can be drawn?

Answer-

There are 20 points on a circle.To draw a chord, 2 points are required.     

∴ the number of chords that can be drawn through 20 points on the circle.

= 20C2                                                        

$=\frac{20!}{{2!18!}$   

$=\frac{20×19×18!}{{2×1×18!}$            

= 190.

10. If 20 points are marked on a circle how many chords can be drawn?

Answer-

There are 20 points on a circle.To draw a chord, 2 points are required.

∴ the number of chords that can be drawn through 20 points on the circle.

= 20C2

$=\frac{20!}{{2!18!}$ 

$=\frac{20×19×18!}{{2×1×18!}$            

= 190.

11.find the number of diagonals of an n sided polygon. In particular, find the number of diagonals when.

a) n=10     b)n=15   c)n=12   d)n =8

Answer-

a)In an n-sided polygon, there are ‘n’ points and ‘n’ sides.               

∴ Through ‘n’ points we can draw nC2 lines including sides.

∴ Number of diagonals in n sided polygon.

= nC2 – n .........(n = number of sides)           

n = 10                                                             

nC2 – n = 10C2 – 10                                

$\frac{10.9}{1.2}-10$                   

= 45 – 10                                                         

= 35

b)There are n vertices in the polygon of n-sides.

 If we join any two vertices, we get either side or the diagonal of the polygon.  

Two vertices can be joined in nC2 ways.      

∴ total number of sides and diagonals = nC2 But there are n sides in the polygon.        

∴ total number of the diagonals = nC2 – n       

n = 15 sides                                                       

∴ the number of diagonal that can be drawn. 

= 15C2 – 15                              

=$\frac{15×14×13!}{2×13!}-15$

=15×142-15

= 105 – 15       

= 90

c)In an n-sided polygon, there are ‘n’ points and ‘n’ sides.

∴ Through ‘n’ points we can draw nC2 lines including sides.      

∴ Number of diagonals in n sided polygon.

= nC2 – n   ...(n = number of sides)

n = 12, 

nC2 – n  = 12C2 – 121 

$=\frac{2×11}{1×2}-12$

= 66 – 12                                                               

= 54

d)There are n vertices in the polygon of n-sides.          

  If we join any two vertices, we get either side or the diagonal of the polygon.

Two vertices can be joined in nC2 ways.          

∴ total number of sides and diagonals = nC2

But there are n sides in the polygon.

∴ total number of the diagonals = nC2 – n 

n = 8 sides                                               

∴ the number of diagonals that can be drawn 

12.There are 20 straight lines in a plane so that no two lines are parallel and no three lines are concurrent. Determine the number of points of intersection.

Answer-

There are 20 lines such that no two of them are parallel and no three of them are concurrent. 

Since no two lines are parallel              

∴ they intersect at a point     

∴ Number of points of intersection if no two lines are parallel and no three lines are concurrent.             

 = 20C2                           

$=\frac{20!}{2!18!}$

$=\frac{20×19×18!}{2×1×18!}$

  = 190 

13.10 points are plotted on a plane. Find the number of straight lines obtained by joining these points if   

A)no three points are collinear.

B) four points are collinear.

Answer-

A )line is drawn by joining 2 points from the given 10 points.    

  ∴ number of straight lines= 10C2                                          

 =$\frac{10!}{2!8!}$

$=\frac{10×9×8!}{2×8!}$

= 45, if no three points are collinear. 

B) when four points are collinear then the total number of lines= 10C2 but it contains lines using that 4 collinear points make only one line

lines using 4 points = 4C2

No. of ways is 4 points are collinear = 10C2 - 4C2 +1

= 45 -  $\frac{4!}{2!2!}$+1

=46- 6

=40 

14.Find the number of triangles formed by joining 12 points if       a) no three points are collinear

b) 4 points are collinear.

Answer-

a)A triangle can be formed by selecting 3 non-collinear points

3 points be selected from 12 points

$^{12}C_3=\frac{12!}{9!3!}$

$=\frac{12.11.10.9!}{9!3!}$

=220 ways

b)There are 12 points on the plane

when 4 of these points are collinear

$\therefore$ number of triangles that can be from these points=$^{12}C_3-^{4}C_3$

=220-$\frac{4!}{3!1!}$

=220-4 

=216

15. A word has three consonant 8 vowels. How many distinct words can be formed if 4 consonants and 2 vowels are use and?

Answer-

4 consonants can be selected from 8 consonants in $^8C_4$ ways 2 vowels can be selected from 3 vowels in $^3C_2$

$\therefore$ number of words with 4 consonants and 2 vowels

=$^8C_4\times ^3C_2$

=$\frac{8!}{4!4!}\times \frac{3!}{2!1!}$

=70 $\times$ 3

=210

now, each of these 6 letters can be arranged in 6P6 ways=6!

$\therefore$ total numbers can be formed 4 consonants and 2 vowels

$=210 \times 6!$

=151200 

16. Find n if,

(i) ^nC_8=^nC_{12}

Answer-

$^nC_{12}=^nC_8$ $^nC_x=^nC_y$ then  either  x=y or  x=n-y 

\therefore 12=n-8    \hspace{2cm} \because 12\neq 8 

n=20

(ii)$^{23}C_{3n}=^{23}C_{2n+1}$

Answer-

$^{23}C_{3n}=^{23}C_{2n+3}$ \because $^nC_x=^nC_y$  then  either  x=y  or  x=n-y

\therefore 3n=2n+3  or 3n+2n+3=23                                        

  $n=3$ or $5n=20$              

\therefore  $n=3$ or $n=4$

(iii)$^{21}C_{6n}=^{21}C_{n^2-5}$

Answer- 

$^{21}C_{6n}=^{21}C_{n^2-5}$                                              

$\because$ $^nC_x=^nC_y$ then either $x=y or x=n-y$      

$\therefore$   $6n=n^2+5$    or     $6n=21-(n^2+5)$

$\therefore$   $n^2-6n+5=0$   or     $6n=21-n^2-5$

$\therefore$   $(n-1)(n-5)=0$   or     $n^2+6n-16=0$

$\therefore$   n=1 or n=5     or     n=-8  or n=2     

if n=5 then $n^2+5=30>21$   or    $n\neq-8 \\ so, n=2$\\               

(iv)^{2n}C_{r-1}=^{2n}C_{r+1}

Answer-

$^{2n}C_{r-1}=^{2n}C_{r+1}$                                              

$\therefore$  $^nC_x=^nC_y$ then either x=y or x=n-y                  

\therefore  r-1=r+1     or        r-1=2n-r-1                              

\therefore   r=1+1      or        r-1+r+1=2n                             

\therefore     r=2      or            r=n

(v) $^{n}C_{n-2}=15$

Answer-

$^{n}C_{n-2}=15$

$\frac{n!}{(n-2)!(n-n+2)!}$=15   

$\frac{n(n-1)(n-2)!}{(n-2)!2!}$=15                                      

$\frac{n(n-1)}{2}=15$                                              

 n(n-1)=15.2                                  

 n(n-1)=30   

 n(n-1)=6.5

  so, n=6

17. Find x if $^{n}P_{r}=x$ \times ^nC_r 

Answer-

$^{n}P_{r}=x$ \times $^nC_r$    

$\frac{n!}{(n-r)!}=x$ \times  $\frac{n!}{r!(n-r)!}$

$1=x \times \frac{1}{r!}$

 x=r! 

18. Find r if

$^{11}C_4+^{11}C_5+^{12}C_6+^{13}C_7=$^{14}C_r$

Answer-   

$^{11}C_4+^{11}C_5+^{12}C_6+^{13}C_7=^{14}C_r$

  As,  $^nC_r+^nC_{r-1}=^{n+1}C_r$

 $^{14}C_r$=$^{11}C_4+^{11}C_5+^{12}C_6+^{13}C_7$

 $=^{11+1}C_5+^{12}C_6+^{13}C_7$

 $=^{12}C_5+^{12}C_6+^{13}C_7$

 $=^{12+1}C_6+^{13}C_7$

 $=^{13}C_6+^{13}C_7$

 $=^{14}C_7$  

  so, r=7

  19. Find the value of $\sum_{r=1}^{4}{} ^{21-r}C_4$\\

Answer- 

 $^{21-r}C_4$ 

$=^{21-1}C_4 + ^{21-2}C_4+ ^{21-3}C_4+^{21-4}C_4$

$=^{20}C_4+^{19}C_4+{18}C_4+{17}C_4$

  $\because$  $^nC_r+^nC_{r-1}$=$^{n+1}C_r$ 

$^nC_{r-1}=^{n+1}C_r-^nC_r$

$=^{20}C_4+^{19}C_4+^{18}C_4+{18}C_5-{17}C_5$

$=^{20}C_4+\left(^{19}C_4+^{19}C_5\right)-{17}C_5$

$=^{20}C_4+^{19+1}C_5-{17}C_5$

$=^{20}C_4+^{20}C_5-{17}C_5$

$=^{21}C_5-{17}C_5$

 $=\frac{21!}{5!(21-5)!}-\frac{17!}{5!(17-5)!}$

 $=\frac{21!}{5!16!}$-$\frac{17!}{5!12!}$

 $=\frac{21.20.19.18.17.16!}{5!16!}$-$\frac{17.16.15.14.13.12!}{5!12!}$

 $=\frac{21.20.19.18.17}{5.4.3.2.1}$-$\frac{17.16.15.14.13}{5.4.3.2.1}$

  = 20349 - 6188                                                              

  =1416                                           

20. Find the differences between the greatest values in the following:

(a) $^{14}C_r and ^{12}C_r

Answer-

here n=14 is an even 

$^{14}C_r$ then the greatest value of nCr occurs at $r=\frac{n}{2}$

$r=\frac{14}{2}$

r=7

Greatest value of $^{14}C_r=^{14}C_7$

$=\frac{14!}{(14-7)!}$

$=\frac{14.13.12.11.10.9.8.7!}{7!(14-7)!}$

$=\frac{14.13.12.11.10.9.8}{7.6.5.4.3.2.1}$

=3432

Also, for the greatest values of ^{12}C_r $

here, n=12 $r=\frac{12}{2}=6$

$^{12}C_6$

$=\frac{12!}{6!(12-6)!}$

$=\frac{12.11.10.9.8.7.6!}{6!6!}$

$=\frac{12.11.10.9.8.7}{6.5.4.3.2.1}$

=924

the difference between 

=$^{14}C_r$ and $^{12}C_r$

=3432-964

=2508

(b)$^{13}C_r$ and $^{8}C_r$

Answer-

the greatest value of

 $^nC_r$ $r=\frac{n}{2}$ if n is even and if n is odd $r=\frac{n-1}{2}$ 

$^{13}C_r=$

here n=13 $r=\frac{13-1}{2}$

$r=\frac{12}{2}=6$

$^{13}C_r={13}C_6$

=$\frac{13!}{6!7!}$

=$\frac{13.12.11.10.9.8.7!}{7!6.5.4.3.2.1}$

=1716

so,$^8C_r$ here n=8

n=\frac{8}{2}=4

$^8C_4$

$=\frac{8!}{4!6!}$

$=\frac{8.7.6.5.4!}{4!6!}$

=70

$\therefore$ the difference between

=$^{13}C_r$ and $^{8}C_r$

=1716-70

=1646 

(c)$^{15}C_r$ and $^{11}C_r$

Answer-

the greatest value of

 $^nC_r$ $r=\frac{n}{2}$ if n is even and if n is odd $r=\frac{n-1}{2}$ 

$^{15}C_r$

here, n=15 so $r=\frac{n-1}{2}=\frac{15-1}{2}=7$

$^{15}C_r$=$^{15}C_7$

$=\frac{15!}{7!8!}$

$=\frac{15.14.13.12.11.10.9.8!}{8!.7.6.5.4.3.2.1}$

=6435

$^{11}C_r$

here, n=11 so $r=\frac{11-1}{2}=5$

$^{11}C_r$=$^{11}C_5$=$\frac{11!}{5!6!}$

$=\frac{11.10.9.8.7.6!}{6!5.4.3.2.1}$

=462

the difference between $^{15}C_r$ and $^{11}C_r$ 

=6435-462

=5973 

21.In how many ways can a boy invite his five friends to a party so that at least 3 join the party?

Answer-

Number of friends n=5

Atleast 3 friends join party

$^5C_3+^5C_4+^5C_5$

$=\frac{5!}{3!(5-3)!}+\frac{5!}{4!}+1$

$=10+5+1$

=16 

22. A group consists of 9 men and 6 women.A team of 6 is to be selected.How many possible selections will have atleast three women.

Answer-

There are 9 men and 6 women

will have 3 women atleast

=3W3M+4W2M+5W1M+6W no men

=$^6C_3 \times ^9C_3+^6C_4 \times ^9C_2+^6C_5 \times ^9C_1+1$

=1680+540+54+1

=2275

Therefore, 2275 can be formed if the team consist atleast 3 women 

23. A committee of 10 person is to be formed from a group 10 women 8 men.How many possible committees will have at least five women? How many possible committee will have men in majority?

Answer-

There are 10 women and 8 men.

A committee of 10 persons is to be formed.If atleast 5 women have to be selected in a committee

so, possible selection are

5 Women 5 Men          6 Women 4 Men

7 Women 3 Men          8 Women 2 Men

9 Women 1 Men          10 Women 0 Men

$\therefore$ the number of ways of forming committees such that at least 5 women are included

$={10}C_5 \times 8C_5+{10}C_6 \times 8C_4+{10}C_7 \times 8C_3+{10}C_8 \times 8C_2+{10}C_9 \times 8C_1+{10}C_{10}\times 8C_0$

=252.56+210.70+120.56+45.28+80+1

=36873

Men in Majority

men in majority means 6,7,8

=$^8C_6 \times ^{10}C_4+^8_7 \times ^{10}C_3+^8C_8 \times ^{10}C_2$

=5880+960+45

=6885 

24. A question paper has two sections, section 1 has 5 questions and section 2 has 6 questions a student must answer at least two questions from each section among six question hi answers. How many different choices does the student have in choosing questions?

Answer-

There are 11 question out of which section 1 has 5 question and section 2 has 6 section\\

possible choices S I-2 S II-4, S I-3 S II-3 S I-4 S II-2

$^5C_2 \times ^6C_4+^5_3 \times ^6C_3+^5C_4 \times ^6C_2$

=$10 \times 15+10 \times 20+5 \times 15$

=150+200+75

=425

$\therefore$ in 425 ways student can select 6 questions taking atleast 2 question from each

25.There are 5 wicket keepers and 5 bowlers among 22 cricket players. A team of 11 players is to be selected so that there is exactly one wicket keeper and atleast 4 bowlers in the team how many different teams can be formed.

Answer-

Among 22 players there are 3 wicketkeepers+5 bowlers=8players and the remaining 22-8=14 players batsmen.

i)1 wicketkeeper 4 bowlers 6 players

ii)1 wicketplayer 5 bowler 5 players

now, number of selection in 

(i)$^3C_1 \times ^5C_4 \times {14}C_6$+$^3C_1 \times ^5C_5 \times {14}C_5$

$=3 \frac{5!}{4!} \frac{14!}{6!9!}+3 \frac{14!}{5!9!}$

=45045+6006

=51051

26.5 students are selected from 11 how many ways can the students be selected if\\

A) two specified students are selected?      

B) to specified student are not selected?

Answer-

A)5 students are to be selected from 11 students

when 2 specified students are always together

then remaining student can be selected from 11-2=9 students

$\therefore$ number of ways of selecting 3 students from 9 students

= $^9C_3$

=$\frac{9!}{3!6!}$

=84

$\therefore$ selection is done in 84 ways when 2 specified students are included

B)Number of students is 11 we have to select 5 students

2 particular student are not selected

if two particular student are not selected then from remaining (11-2)=9 students, we have to select 5 students

This can be done^9C_5$ ways

$\therefore$ total number of selections

$^9C_5$

$=\frac{9!}{5!4!}$

$=\frac{9.8.7.6.5!}{5!4!}$

=196

                                

Thanks for reading inlogical math

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