How do you solve #\frac { 2} { 5! ( n - 5) ! } - \frac { 1} { 4! ( n - 4) ! } - \frac { 1} { 6! ( n - 6) ! } = 0#?

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Answer 1 · 2017-05-21T16:41:44

#n = {7,14}#

#(n-5)! = (n-6)!(n-5)# and
#(n-4)! = (n-6)!(n-5)(n-4)# so

#\frac { 2} { 5! ( n - 5) ! } - \frac { 1} { 4! ( n - 4) ! } - \frac { 1} { 6! ( n - 6) ! } = (2/(5!(n-5))-1/(4!(n-5)(n-4))-1/(6!))1/((n-6)!)=0#

then

#2/(5!(n-5))-1/(4!(n-5)(n-4))-1/(6!)=0#

Now supposing #n ne 4# and #n ne 5#

#(2(n-4))/(5!)-1/(4!)-((n-4)(n-5))/(6!)=0#

solving for #n# we get #n = {7,14}#