DJ Cynthia is making a playlist for a radio show; she is trying to decide what 10 songs to play and in what order they should be played. If she has her choices narrowed down to 8 blues, 4 rock, 4 jazz, and 7 pop songs, and she wants to play no more than 4 pop songs, how many different playlists are possible?

3,770,461,100,000 different playlists are possible.

Step-by-step explanation:

To solve this question, the arrangements formula and the combinations formula are used. The arrangements formula is used considering the order of the musics being played, and the combinations formula is used to find the number of possible combinations of musics.

Arrangements of n elements:

The number of possible arrangements of n elements is given by:

Combinations formula:

is the number of different combinations of x objects from a set of n elements, given by the following formula.

0 pop songs

10 non-pop from a set of 8 + 4 + 4 = 16.

Arrangements of 10(order in which the musics are played). So

1 pop song:

9 non-pop from a set of 16, 1 pop from a set of 7. So

2 pop songs:

8 non-pop from a set of 16, 2 pop from a set of 7. So

3 pop songs:

7 non-pop from a set of 16, 3 pop from a set of 7. So

4 pop songs:

6 non-pop from a set of 16, 4 pop from a set of 7. So

How many different playlists are possible?

3,770,461,100,000 different playlists are possible.

  5.   y=6; de=7; fg=8

  6.   x=4.2; jl=14.2; mn=13

step-by-step explanation:

when chords cross inside a circle, the product of the segment lengths is a constant.

5. dh×eh = fh×gh

  y = gh = dh×eh/fh = 3×4/2 . . . divide by the coefficient of gh

  y = 6

of course, the total chord length is the sum of the lengths of its segments.

  de = dh +eh = 3+4 = 7

  fg = fh +gh = 2+6 = 8

6. this problem works the same way as the previous one.

  x = lk = mk×nk/jk = 6×7/10 = 4.2

  jl = jk +lk = 10 +4.2 = 14.2

  mn = mk +nk = 6 +7 = 13

comment on chords and secants

you can think of the points h (problem 5) and k (problem 6) as points where the chords meet. a chord is part of a secant line, a line that intersects the circle in two points. one can also consider these points (h or k) to be the points where the respective secant lines meet.

then, the product that is a constant is the product of the distance from the meeting point (h or k) to one circle intersection with the distance from that meeting point to the other circle intersection. for problem 5, the constant is the product hd×he = hf×hg.

it turns out that this rule regarding the products of lengths to points of intersection is also true when the secants intersect outside the circle.

that is, you only need to remember one rule to work both kinds of problems.

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