SolutionnP8+nP15+nP20+1=n(equation1)
nP9+nP13+4=n(equation2)
equation1=equation2
nP8+nP15+nP20+1=nP9+nP13+4
If each term of the equation is rewritten using factorial function, then the equation becomes
(n−8)!n!+(n−13)!n!+(n−20)!n!+1=(n−9)!n!+(n−15)!n!+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)=(20−8)!20!+(20−13)!20!+(20−20)!20!+1−(20−9)!20!−(20−15)!20!−4=2.41∗1018
f(21)=(21−8)!21!+(21−13)!21!+(21−20)!21!+1−(21−9)!21!−(21−15)!21!−4=5.1∗1019
f(22)=(22−8)!22!+(22−13)!22!+(22−20)!22!+1−(22−9)!22!−(22−15)!22!−4=5.62∗1020
f(23)=(23−8)!23!+(23−13)!23!+(23−20)!23!+1−(23−9)!23!−(23−15)!23!−4=4.31∗1021
f(24)=(24−8)!24!+(24−13)!24!+(24−20)!24!+1−(24−9)!24!−(24−15)!24!−4=2.59∗1022
f(25)=(25−8)!25!+(25−13)!25!+(25−20)!25!+1−(25−9)!25!−(25−15)!25!−4=1.29∗1023
f(26)=(26−8)!26!+(26−13)!26!+(26−20)!26!+1−(26−9)!26!−(26−15)!26!−4=5.6∗1023
f(27)=(27−8)!27!+(27−13)!27!+(27−20)!27!+1−(27−9)!27!−(27−15)!27!−4=2.16∗1024
f(28)=(28−8)!28!+(28−13)!28!+(28−20)!28!+1−(28−9)!28!−(28−15)!28!−4=7.56∗1024
∴ 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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