6 horses {A,B,C,D,E,F} Participate in a race. If there are no draws in

Question

6 horses {A,B,C,D,E,F} Participate in a race. If there are no draws in the rate then in how many ways can the race end such that horse A always finishes ahead of C and B both the horses?

A) 720
B) 360
C) 240
D) 120
E) 48

Source: https://www.GMATinsight.com

goodie · Answer

There are four cases:

#1 Horse A finished first (B and C were put arbitrary):

A  _ (B) _ (C) _ = all other houses can be arranged in 5! ways = 120

#2 Horse A finished second (and etc.):

_  A  _ (B) _ (C) = 3C1 for the first place and 2C2*4! For the last 4 places = 3*1*24 = 72

#2 third:

_ _  A  _ (B) (C) = 3C2 * 3! = 36

And finally A finished fourth:

_ _ _  A  (B) (C) = 3C3*3!*2! = 12

Total = 120 + 72 + 36 + 12 = 240

Answer C

subsauce · Answer

This reminds me of another problem I've seen where instead of A always finishing infront of both B and C, A always finished infront of Just B.

In this situation there are 6! combinations but 1/2 of the total possibilities include B finishing in front of A. Thus the solution would be 6!/2 = 360

In this case, A finishes in front of both B and C. There are equal possibilities that B finishes in front of A and C, and that C finishes in front of A and B. Thus A will finish in front of BOTH B and C 1/3 of the time.

So the answer logically is 1/3 of the total possibilities.

6! / 3 = 240 which is the correct answer.

GMATinsight · Answer

METHOD 1:

We need 3 places for A B and C which can be selected in 6C3 ways

Now A has to seated on the first selected place and B and C can exchange position on the remaining two positions in 2! ways

Remaining three individuals can occupy the positions in 3! ways

Total ways of arranging them as desired = 6C3 * 2! * 3! = 20*2*6 = 240

METHOD 2:

A has equal chances of being ahead of B and C as B and C have to be ahead of other two hence

probability of A to be ahead of both B and C = 1/3

Total ways of arranging the six individuals = 6! = 720 ways

Favourable cases = (1/3)*720 = 240

METHOD 3:    
If A comes at first place A - - - - - then arrangements = 5!
If A comes at Second place - A - - - - then arrangements = 4C2*2!*3! = 72
If A comes at Third place - - A - - - then arrangements = 3C2*2!*3! = 36
If A comes at Forth place - - - A - - then arrangements = 2C2*2!*3! = 12

Total ways = 120+72+36+12 = 240

Answer: Option C

satya2029 · Answer

Condition is on A, B and C.
Take them out. We are left with D,E and F
 A comes at 1st place- 1 way and remaining can be arranged in !5 ways 
so total=1*!5=120
 A comes at 2nd place- First place can be taken by D,E and F in 3 ways, for 2nd place only 1 way, then remaining 4 places in !4 ways 
So total=3*1*!4=72
 A comes at 3rd place- First place can be taken by D,E and F in 3 ways, 2nd place by again two of D,E and F in 2 ways, for 3rd place 1 way ,for remaining places !3 ways 
so total=3*2*1*!3=36
 A comes at 4th place- First three places taken by D,E and F in !3 ways, fourth place 1 way, remaining 2 places in !2 ways 
So total=!3*1*!2=12
Total=12+72+36+12=240
C:)

bumpbot · Answer

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