# Goldenrod and No Hope are in a horse race with 6 contestants. How many different arrangements of finishes are there if No Hope always finishes before Goldenrod and if all of the horses finish the race?

A. 700

B. 360

C. 120

D. 24

E. 21

**Answer: Option B **

__Solution(By Examveda Team)__

Two horses A and B, in a race of 6 horses . . . A has to finish before B.If A finishes 1 . . . . . B could be in any of other 5 positions in 5 ways and other horses finish in 4! Ways, so total ways = 5 × 4!

If A finishes 2 . . . . . B could be in any of the last 4 positions in 4 ways. But the other positions could be filled in 4! ways, so the total ways = 4 × 4!

If A finishes 3

^{rd}. . . . . B could be in any of last 3 positions in 3 ways, but the other positions could be filled in 4! ways, so total ways = 3 × 4!

If A finishes 4

^{th}. . . . . B could be in any of last 2 positions in 2 ways, but the other positions could be filled in 4! ways, so total ways = 2 × 4!

If A finishes 5

^{th}. . . . B has to be 6th and the top 4 positions could be filled in 4! ways.

A cannot finish 6

^{th}, since he has to be ahead of B.

Therefore total number of ways:

= 5 × 4! + 4 × 4! + 3 × 4! + 2 × 4! + 4!

= 120 + 96 + 72 + 48 + 24

= 360

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