Question

5.
If nPr= 1814400 and nCr = 45, find n+4Cr+3​

Answer 1

Answer:

Can not be determined.

Step-by-step explanation:

Given,

$^nP_r = 1814400$

$\implies \frac{n!}{(n-r)!}=1814400----(1)$

$^nC_r = 45$

$\implies \frac{n!}{r!(n-r)!}=45----(2)$

$\frac{Equation (1)}{Equation (2)}$

We get,

$\frac{1}{1/r!}=\frac{181440}{45}$

$r!=4032$

$r(r-1)(r-2)...(3)(2)(1) = 4032$

$\frac{r+1}{2}\times r = 4032$

$r^2 + r = 8064$

By the quadratic formula,

r = -90.301 or 89.301

Since, value of r can neither negative number nor a decimal number.

Hence, the value of $^{n+4}C_{r+3}$ CAN NOT BE DETERMINED.