Answer to Question #128646 in Combinatorics | Number Theory for Hussain Ousman Darboe

Question

nP8+nP15+nP20+1=n (equation 1)
nP9+nP13+4=n (equation 2)

Answer 1

$$Solution$$

$nP8+nP15+nP20+1=n (equation 1)$

$nP9+nP13+4=n (equation 2)$

$equation1=equation2$

$nP8+nP15+nP20+1=nP9+nP13+4$

If each term of the equation is rewritten using factorial function, then the equation becomes

$$\frac{n!}{(n−8)!}​+\frac{n!}{(n−13)!}​+\frac{n!}{(n−20)!}+1=\frac{n!}{(n−9)!}​+\frac{n!}{(n−15)!}​+4$$

Now to solve for the value of n, we need a function f(n);

$f(n) = nP8 + nP13 + nP20 + 1- nP9 - nP15 - 4$

Let's find the value of n that satisfies the equation

For n = 20, 21, 22, 23, 24, 25, 26, 27, 28 since f(n) is a monotonically increasing function of n.

$f(20) =\frac{20!}{(20−8)!}​+\frac{20!}{(20−13)!}​+\frac{20!}{(20−20)!}+1-\frac{20!}{(20−9)!}​-\frac{20!}{(20−15)!}​-4= 2.41 * 10^{18}$

$f(21) =\frac{21!}{(21−8)!}​+\frac{21!}{(21−13)!}​+\frac{21!}{(21−20)!}+1-\frac{21!}{(21−9)!}​-\frac{21!}{(21−15)!}​-4= 5.1 * 10^{19}$

$f(22) =\frac{22!}{(22−8)!}​+\frac{22!}{(22−13)!}​+\frac{22!}{(22−20)!}+1-\frac{22!}{(22−9)!}​-\frac{22!}{(22−15)!}​-4= 5.62 * 10^{20}$

$f(23) =\frac{23!}{(23−8)!}​+\frac{23!}{(23−13)!}​+\frac{23!}{(23−20)!}+1-\frac{23!}{(23−9)!}​-\frac{23!}{(23−15)!}​-4= 4.31 * 10^{21}$

$f(24) =\frac{24!}{(24−8)!}​+\frac{24!}{(24−13)!}​+\frac{24!}{(24−20)!}+1-\frac{24!}{(24−9)!}​-\frac{24!}{(24−15)!}​-4= 2.59 * 10^{22}$

$f(25) =\frac{25!}{(25−8)!}​+\frac{25!}{(25−13)!}​+\frac{25!}{(25−20)!}+1-\frac{25!}{(25−9)!}​-\frac{25!}{(25−15)!}​-4= 1.29 * 10^{23}$

$f(26) =\frac{26!}{(26−8)!}​+\frac{26!}{(26−13)!}​+\frac{26!}{(26−20)!}+1-\frac{26!}{(26−9)!}​-\frac{26!}{(26−15)!}​-4= 5.6 * 10^{23}$

$f(27) =\frac{27!}{(27−8)!}​+\frac{27!}{(27−13)!}​+\frac{27!}{(27−20)!}+1-\frac{27!}{(27−9)!}​-\frac{27!}{(27−15)!}​-4= 2.16 * 10^{24}$

$f(28) =\frac{28!}{(28−8)!}​+\frac{28!}{(28−13)!}​+\frac{28!}{(28−20)!}+1-\frac{28!}{(28−9)!}​-\frac{28!}{(28−15)!}​-4= 7.56 * 10^{24}$

$\therefore$ Note that the equation $f(n)$ does not hold for any value of $n$ .

Here is the graph of the function.

Notice that the value of $f(n)$ is varying abruptly.

From the graph, we can clearly see a monotonically increasing trend.

We therefore conclude that the equation does not hold for any value of n.