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
SolutionSolution

nP8+nP15+nP20+1=n(equation1)nP8+nP15+nP20+1=n (equation 1)

nP9+nP13+4=n(equation2)nP9+nP13+4=n (equation 2)


equation1=equation2equation1=equation2

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


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


n!(n8)!+n!(n13)!+n!(n20)!+1=n!(n9)!+n!(n15)!+4\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+1nP9nP154f(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)=20!(208)!+20!(2013)!+20!(2020)!+120!(209)!20!(2015)!4=2.411018f(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)=21!(218)!+21!(2113)!+21!(2120)!+121!(219)!21!(2115)!4=5.11019f(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)=22!(228)!+22!(2213)!+22!(2220)!+122!(229)!22!(2215)!4=5.621020f(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)=23!(238)!+23!(2313)!+23!(2320)!+123!(239)!23!(2315)!4=4.311021f(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)=24!(248)!+24!(2413)!+24!(2420)!+124!(249)!24!(2415)!4=2.591022f(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)=25!(258)!+25!(2513)!+25!(2520)!+125!(259)!25!(2515)!4=1.291023f(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)=26!(268)!+26!(2613)!+26!(2620)!+126!(269)!26!(2615)!4=5.61023f(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)=27!(278)!+27!(2713)!+27!(2720)!+127!(279)!27!(2715)!4=2.161024f(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)=28!(288)!+28!(2813)!+28!(2820)!+128!(289)!28!(2815)!4=7.561024f(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)f(n) does not hold for any value of nn .


Here is the graph of the function.




Notice that the value of f(n)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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