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

Question #128646
nP8+nP15+nP20+1=n (equation 1)
nP9+nP13+4=n (equation 2)
1
Expert's answer
2020-08-09T18:17:31-0400
"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\u22128)!}\u200b+\\frac{n!}{(n\u221213)!}\u200b+\\frac{n!}{(n\u221220)!}+1=\\frac{n!}{(n\u22129)!}\u200b+\\frac{n!}{(n\u221215)!}\u200b+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\u22128)!}\u200b+\\frac{20!}{(20\u221213)!}\u200b+\\frac{20!}{(20\u221220)!}+1-\\frac{20!}{(20\u22129)!}\u200b-\\frac{20!}{(20\u221215)!}\u200b-4= 2.41 * 10^{18}"


"f(21) =\\frac{21!}{(21\u22128)!}\u200b+\\frac{21!}{(21\u221213)!}\u200b+\\frac{21!}{(21\u221220)!}+1-\\frac{21!}{(21\u22129)!}\u200b-\\frac{21!}{(21\u221215)!}\u200b-4= 5.1 * 10^{19}"


"f(22) =\\frac{22!}{(22\u22128)!}\u200b+\\frac{22!}{(22\u221213)!}\u200b+\\frac{22!}{(22\u221220)!}+1-\\frac{22!}{(22\u22129)!}\u200b-\\frac{22!}{(22\u221215)!}\u200b-4= 5.62 * 10^{20}"


"f(23) =\\frac{23!}{(23\u22128)!}\u200b+\\frac{23!}{(23\u221213)!}\u200b+\\frac{23!}{(23\u221220)!}+1-\\frac{23!}{(23\u22129)!}\u200b-\\frac{23!}{(23\u221215)!}\u200b-4= 4.31 * 10^{21}"


"f(24) =\\frac{24!}{(24\u22128)!}\u200b+\\frac{24!}{(24\u221213)!}\u200b+\\frac{24!}{(24\u221220)!}+1-\\frac{24!}{(24\u22129)!}\u200b-\\frac{24!}{(24\u221215)!}\u200b-4= 2.59 * 10^{22}"


"f(25) =\\frac{25!}{(25\u22128)!}\u200b+\\frac{25!}{(25\u221213)!}\u200b+\\frac{25!}{(25\u221220)!}+1-\\frac{25!}{(25\u22129)!}\u200b-\\frac{25!}{(25\u221215)!}\u200b-4= 1.29 * 10^{23}"


"f(26) =\\frac{26!}{(26\u22128)!}\u200b+\\frac{26!}{(26\u221213)!}\u200b+\\frac{26!}{(26\u221220)!}+1-\\frac{26!}{(26\u22129)!}\u200b-\\frac{26!}{(26\u221215)!}\u200b-4= 5.6 * 10^{23}"


"f(27) =\\frac{27!}{(27\u22128)!}\u200b+\\frac{27!}{(27\u221213)!}\u200b+\\frac{27!}{(27\u221220)!}+1-\\frac{27!}{(27\u22129)!}\u200b-\\frac{27!}{(27\u221215)!}\u200b-4= 2.16 * 10^{24}"


"f(28) =\\frac{28!}{(28\u22128)!}\u200b+\\frac{28!}{(28\u221213)!}\u200b+\\frac{28!}{(28\u221220)!}+1-\\frac{28!}{(28\u22129)!}\u200b-\\frac{28!}{(28\u221215)!}\u200b-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.




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