Exercise 3.6 3) 4) 5)

 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

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

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

$^{14}C_{11}=364$ 

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