combinatorics

[Winner2013]

Q. A pod of 6 dolphins always swims single file, with 3 females at the front and 3
males in the rear. In how many different arrangements can the dolphins swim?

source - MGMAT strategy guide

answer is - 36

I want to know where am i going wrong in my logic. Can someone help?

My logic :

6!
/ 3! * 3!

where am i going wrong?

thanks,

[parveen110]

Q. A pod of 6 dolphins always swims single file, with 3 females at the front and 3
males in the rear. In how many different arrangements can the dolphins swim?

source - MGMAT strategy guide

answer is - 36

I want to know where am i going wrong in my logic. Can someone help?

My logic :

6!
/ 3! * 3!

where am i going wrong?

thanks,

According to the question the arrangement is like FFFMMM

Let's consider FFF as one item and MMM as another.

Now, the three female dolphins, FFF, may be arranged within themselves in 3! ways.
Similarly, the three male dolphins, MMM, may be arranged in 3! ways.

thereby giving, 3!x3!= 36 ways.

Your logic, however, appropriately answers following kind of question:
There are 3 red and 3 green marbles in a bag, they are drawn one by one and arranged in a row, assuming that all 6 marbles are drawn, determine the number of different arrangements.

Logic involved here is similar to what you have used, i.e, permutation of n things NOT all different.

Hope, i made sense:)

[GMATGuruNY]

Q. A pod of 6 dolphins always swims single file, with 3 females at the front and 3
males in the rear. In how many different arrangements can the dolphins swim?

source - MGMAT strategy guide

answer is - 36

I want to know where am i going wrong in my logic. Can someone help?

My logic :

6!
/ 3! * 3!

where am i going wrong?

thanks,

Your solution would be correct for the following problem:
How many ways can the letters AAABBB be arranged?
Number of ways to arrange 6 distinct elements = 6!.
But the arrangement here includes 3 identical A's and 3 identical B's.
To account for the identical elements, we must divide by the number of ways each set of identical elements can be arranged.
The reason:
When the identical elements swap places, the arrangement doesn't change.
Thus, the number of ways to arrange AAABBB = 6!/(3!3!) = 20.
We divide by 3! to account for the 3 identical A's and by another 3! to account for the 3 identical B's.

In the posted problem:
The 3 female dolphins must occupy the 3 front positions, while the 3 male dolphins must occupy the 3 back positions.
In the 3 front positions, the number of ways to arrange the 3 female dolphins = 3!.
In the 3 back positions, the number of ways to arrange the 3 male dolphins = 3!.
To combine these options, we multiply:
3! * 3! = 36.
Since the dolphins are not identical, no division is necessary.

[Jeff@TargetTestPrep]

Q. A pod of 6 dolphins always swims single file, with 3 females at the front and 3
males in the rear. In how many different arrangements can the dolphins swim?

A. 20
B. 36
C. 40
D. 18
E. 54

We are given that a pod of 6 dolphins always swims single file, with 3 females at the front and 3 males in the rear. Thus:

The number of ways to arrange the 3 female dolphins in the front is 3! = 3 x 2 x 1 = 6.

The number of ways to arrange the 3 male dolphins is in the rear is 3! = 3 x 2 x 1 = 6.

Thus, the number of ways to arrange all the dolphins is 6 x 6 = 36 ways.

Answer: B