# help on math SAT problem.

<p>ok, here are two math problems that I could not get correct... dont know why.
The acme plumbing company will send a team of 3 plumbers to work on a certain job. The company has 4 experienced plumbers and 4 trainees. If a team consists of 1 experienced plumber and 2 trainees, how many different such teams are possible.
I keep getting: 4x4x3=48... but the answer is 24</p>

<p>Also, at time t=0, a ball is thrown upwards from an initial height of 6 feet. Until the ball hit the ground, its height, in feet, after t seconds was given by the function h (h=c-(d-4t)^2) in which c and d are positive constants. If the ball reached its max height of 106 feet at time t=2.5 seconds, what was the height, in feet, of the ball at time t=1.
Thanks guys!</p>

<p>The first one is (4C1)(4C2)=24</p>

<p>At time t=0, a ball is thrown upwards from an initial height of 6 feet. Until the ball hit the ground, its height, in feet, after t seconds was given by the function h (h=c-(d-4t)^2) in which c and d are positive constants. If the ball reached its max height of 106 feet at time t=2.5 seconds, what was the height, in feet, of the ball at time t=1.</p>

<p>Use the first data point (t=0, h=6) to get 6=c - d^2 or c=6+d^2
Use the second data point (t=2.5, h=106) to get 106=c - (d-10)^2 ... the 10 comes from 4t=4<em>2.5=10. These two equations can be solved to give d=10 and c=106. The easiest way to solve them is to substitute c=6+d^2 in the second equality and recall that x^2-y^2=(x+y)</em>(x-y)</p>

<p>You want the height when t=1. It is c-(d-4)^2 ... the 4 comes from 4t=4*1. So substitute for c and d, from above, to get 106-(10-4)^2 which is 70.</p>

<p>It this a SAT 2 question? It would never appear in SAT 1.</p>

<p>@clanddarkfire- Thanks. Actually found out that I was doing permutations haha.
@ fogcity- Thanks. And yes, this was an SAT 1 question. Pretty tough for an SAT 1 q if u ask me lol</p>

<p>wait, how do you know when to use permutation or combinations? I forgot</p>

<p>combinations- order doesnt matter
permutations- order matters</p>