hihihi

\(C^n_r=\frac{n!}{r!(n-r)!}\)

\(C^6_4=\frac{6!}{4!(6-2)!}\)

\(C^6_4=\frac{6!}{4!*2!}\)

\(C^6_4=\frac{6*5}{2!}\)

\(C^6_4=\frac{30}{2}\)

\(C^6_4=15\)

\(\boxed{15} \)

Another way to see this  =  total arrangements possible  less the arrangements where the math books are together

Total arrangements  =  6!   = 720

Total arrangemens where the  math books are together = they can occupy any of 5 positions....and for each of these positions, they can be arranged in two ways......and for each of these, the other books can be arranged in 4! ways  ......so we have     5 * 2! * 4!  =  10 * 24   = 240

So........720 −  240   =    480 ways

3(3x+y=1/3)

2x-3y=8/3

11x=11/3

\(x=1/3, y=-2/3\)

Ok so,

Let the number of hours after 2:00 p.m. that the two engineers arrive at Starbucks be x and y, and let the number of hours after 2:00 p.m. that the boss arrives at Starbucks be z. Then \(0\le x,y,z\le2\) and in three dimensions, we are choosing a random point within this cube with volume 8. We must have \(z>x\) and \(z>y\); this forms a square pyramid with base area 4 and height 2, or volume 8/3.

However, if one of the engineers decides to leave early, the meeting will fail. The engineers will leave early if \(x>y+1\) or \(y>x+1\). The intersections of these with our pyramid gives two smaller triangular pyramids each with base area 1/2 and height 1, or volume 1/6.

In all, the probability of the meeting occurring is the volume of the big square pyramid minus the volumes of the smaller triangular pyramids divided by the volume of the cube: \(\frac{8/3-1/6-1/6}8=\frac{7/3}8=\boxed{7/24}\)

6960

7/24