xample 13 Find r, if 5 4P = 6 5P .

1.	xample 13 Find r, if 5 4P = 6 5P .
Answer» nPr = n!/(n - r)!So, 4Pr = 4!/(4 - r)! and 5P(r - 1) = 5!/(6 - r)! ii) Hence, given equation is: 54!/(4 - r)! = 65!/(6 - r)!==> 5!/(4 - r)! = 6*5!/{(6 - r)(5 - r)(4 - r)!} ==> 1 = 6/(6 - r)(5 - r)==> 6 = (6 - r)(5 - r)==> 6 = 30 - 11r + r²==> r² - 11r + 24 = 0==> (r - 8)(r - 3) = 0So r = 8 or 3 But r = 8 is not possible, since r has to be less than or equal to 4 Thus r = 3 only

Answer»

nPr = n!/(n - r)!So, 4Pr = 4!/(4 - r)! and 5P(r - 1) = 5!/(6 - r)!

ii) Hence, given equation is: 5*4!/(4 - r)! = 6*5!/(6 - r)!==> 5!/(4 - r)! = 6*5!/{(6 - r)(5 - r)(4 - r)!}

==> 1 = 6/(6 - r)(5 - r)==> 6 = (6 - r)(5 - r)==> 6 = 30 - 11r + r²==> r² - 11r + 24 = 0==> (r - 8)(r - 3) = 0So r = 8 or 3

But r = 8 is not possible, since r has to be less than or equal to 4

Thus r = 3 only

Discussion