Alice, Bobby, Cindy, Daren and Eddy participate in a marathon. If each

Question

Alice, Bobby, Cindy, Daren and Eddy participate in a marathon. If each of them finishes the marathon and no two or three athletes finish at the same time, in how many different possible orders can the athletes finish the marathon so that Alice finishes before Bobby and Bobby before Cindy?

A) 18
B) 20
C) 24
D) 30
E) 36

Source - Math Revolution

Please kindly explain your workings

A) 18
B) 20
C) 24
D) 30
E) 36

Source - Math Revolution

Please kindly explain your workings

SirFrancis · Answer

Without any constrains there a 5! combinations,

The constraints removes 2 degrees of freedom so 3!

the answer is 5!/3!= 5*4= 20

chetan2u · Answer

All 5 can be arranged in 5! Ways but without any conditions..
But A,B and C can be arranged in 3! Amongst themselves and ONLY one out of the 3! Will have ABC in that order amongst themselves.
So divide all by 3!  \frac{5!}{3!} =20
B

GMATAspirer09 · Answer

hi chetan,

why do we need to divide by 3! ?
can you explain in more detail please.

thanks!

Pritu · Answer

We need to divide by 3! in order to ensure that ABC order is maintained. 
If we don't divide by 3 ! then ABC will be arranged in 3! ways = BAC, CAB, CBA , etc .AND we don't want this. We just want ABC order. so In order to ensure that arrangement be only in ABC , we need to divide it by 3 !

Novice90 · Answer

Shouldn't the answer be just 6?

Considering "ABC" as one set, where have 3 things to arrange"ABC","D" and "E", which can be arranged in 6 ways

Here are the 6 sets:

D ABC E
E ABC D
D E ABC
E D ABC
ABC D E
ABC E D

Nikkb · Answer

Question stem says , A should be before B and B should be before C. It doesnot says they all come in line.

So possible combinations:

A B C D E
A B C E D
A D B C E
A E B C D
A B D E C
A B E D C
..... and so on.

Total will be 20 possible orders.

Answer: B

Shobhit7 · Answer

Order for slotting A,B,C is fixed i.e.1.

But Order for D&E is not under any constraint and hence is 5P2.

(Position of A-B-C will be moved, retaining their constrained order, to accommodate D&E in any slot from 5P2).

Therefore, Total Possible Orders of arrangements are=
1x5P2 = 20.

Hence, Ans B

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ScottTargetTestPrep · Answer

Solution:

If A, B, and C are all together and in that order, we can treat   as a single entity, and we have EF , FE ,  EF,  FE, E F, or F E. We see that there are 6 such ways.

If A, B, and C are separate, we can place one of E and F between one of the two pairs {A, B} and {B, C} and the other between the other pair. That is, the order of the 5 people could be AEBFC or AFBEC. We see that there are 2 such ways.

If A and B are together and C is separate from A and B, we can place one of E and F between AB and C and the other before A or after C OR both E and F between AB and C. That is, the order of the 5 people could be ABECF, FABEC, ABFCE, EABFC, ABEFC, or ABFEC. We see that there are 6 such ways.

There should also be 6 ways if B and C are together and A is separate from B and C. Therefore, there are a total of 6 + 2 + 6 + 6 = 20 ways.

Alternate Solution:

Let’s suppose that A finished before B and B finished before C. Thus, not considering D or E, the arrangement looks like the following:

_ A _ B _ C _

We see that there are four possible positions for the athlete D. After we place D (say to the first position), the arrangement will look like the following:

_ D _ A _ B _ C _

Now, we see that there are five possible positions for athlete E. Notice that it does not matter where we placed D in the previous step; in addition to the four options that were available for D, E can also be placed right before or right after D, thus the number of options will increase by one. Notice also that any ordering where A is before B and B is before C can be obtained this way by placing D and E in appropriate positions.

Thus, there are 4 x 5 = 20 different orderings where A is ahead of B and B is ahead of C.

Answer: B

KarishmaB · Answer

5 people A, B, C, D and E are arranged in a line. Total number of ways in which they can be arranged = 5! = 120. For example ABCDE, ABDCE, BDAEC etc. (assuming when A is written to the left of B, it means A is before B etc.) In some of these cases, A is before B (in ABCDE), in others it is after B (in BDAEC). We need only those cases in which A is before B and B is before C. If we were to arrange only A, B and C, we could do it in 3! = 6 ways. ABC or BCA or CAB etc. All these cases will equally distributed among the 120 cases above since A, B and C are all equivalent elements. Of these 6 cases, only ABC is acceptable. Hence only 1/6th of the total 120 cases are acceptable for example ABCDE, ADBEC, EADBC are all acceptable but BACDE, CABDE etc are all not acceptable. Hence only 20 cases are acceptable Answer (B) The concept is explained in detail in this blog post: https://anaprep.com/combinatorics-linea ... -symmetry/ Fundamental Counting Principle is discussed here: https://youtu.be/LFnLKx06EMU

samarpan.g28 · Answer

Basically A, B and C will follow a pattern on their order of completion i.e. ABC. D and E can finish according to that.

1. ABC _ _   --> D and E can arrange themselves in 2!=2 ways.
2. AB _ _ C  --> D and E can arrange themselves in 2!=2 ways.
3. A _ _ BC  --> D and E can arrange themselves in 2!=2 ways.
4. _ _  ABC  --> D and E can arrange themselves in 2!=2 ways.
5. _ A _ B _ C _ -->total 4 spaces available out of which, 2 will be occupied by D and E i.e. 4C2 and they can arrange themselves in 2! ways i.e. 4C2*2!=12 ways.

2+2+2+2+12=20 ways. Option (B) is correct.

Alice, Bobby, Cindy, Daren and Eddy participate in a marathon. If each

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Alice, Bobby, Cindy, Daren and Eddy participate in a marathon. If each

Prep Toolkit

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