find n if nP5=42×nP3; n>4
Answers
Answered by
7
nP5=42×nP3
nP5=42×nP3, therefore
n/(n−5)=42×n/(n−3)
...if we cancel out n!n! on both the sides, then we get:
1=42(n−5)(n−4)
... the solution of which leads to a quadratic with relatively large numbers and so on and so forth...on the other hand, if we keep the respective sides of the equation separate and solve them, then we get:
n(n−1)(n−2)(n−3)(n−4)=42.[n(n−1)(n−2)]
...and since n>4, n(n−1)(n−2) is not equal to zero, hence dividing both sides by that yields:
(n−3)(n−4)=42
which gives an answer of n=10
nP5=42×nP3, therefore
n/(n−5)=42×n/(n−3)
...if we cancel out n!n! on both the sides, then we get:
1=42(n−5)(n−4)
... the solution of which leads to a quadratic with relatively large numbers and so on and so forth...on the other hand, if we keep the respective sides of the equation separate and solve them, then we get:
n(n−1)(n−2)(n−3)(n−4)=42.[n(n−1)(n−2)]
...and since n>4, n(n−1)(n−2) is not equal to zero, hence dividing both sides by that yields:
(n−3)(n−4)=42
which gives an answer of n=10
Similar questions