A boat has a crew of eight. Two of those eight can row only on the str

Bunuel chetan2u KarishmaB

Though Alheli has tried to explain the OA, I am not clear why did we use permutation for this ques? Can you please provide alternate solution?

Further, it seems we need to be aware about the basics of boat structure that bow side and stroke side are alternate to each other. Many people including me don't know this concept as apparent from % correct. I hope GMAT would clarify this within ques itself.

I rowed and never heard of the distinction between sides characterized this way. Usually port and starboard, and bow and stroke as front and back.

But it should be clear, regardless, that two different sides are being discussed and that's all that's necessary.

So, the two people are assigned to the "stroke" side and three to the other.

This leaves two people to be selected from the remaining 3 to sit on the stroke side and 1 of the 3 to the other.

2 people can be selected from the 3

3!/2! = 3 ways. The remaining 1 person is then assigned to the other side.

This takes care of who goes on each side.

But the people on each side can then be arranged 4! = 24 ways, and each of those 24 arrangements on a side is associated with 1 arrangement on the other. So the total ways is

3*24*24 = 1728

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Kushchokhani
I also would say some clarity in question on the terms used would make it better.

However, in the present form too where we speak of rowing, one can infer that two sides are there with four rowers rowing on each side.

Ok, Why permutations? Because these 4 people can be arranged on 4 seats in 4! ways.

Solution:

Step 1: Separate out 2 for strike and 3 for bow side. Only 1 way possible.
Step 2: Further we can select one more rower required at Bow side from remaining 8-2-3 or 3 rowers in 3C1 or 3 ways.
Step 3: After we have selected 4 on each side, they can be arranged in 4! ways on each side.

Ways = 1*3*4!*4! = 3*24*24 = 1728

E

A boat has a crew of eight. Two of those eight can row only on the stroke side, while three can row only on the bow side. In how many ways can the two sides of the boat be manned?

A. 12
B. 36
C. 144
D. 288
E. 1728




Two out of the eight can be on the stroke side only and three out of the eight can be on the bow side only. The remaining 3 can be on either side.

Choosing the missing 2 for the the stroke side: 3C2 = 3. Arranging all 4 on the stoke side: 4! = 24. Total for the stroke side = 3*24 = 72.
The remining 1 out of 3, will automatically be left for the bow side. Arranging 4 on the bow side: 4! = 24. Total for the bow side = 1*24 = 24.

Grand total = 72*24 = 1728.

Answer: E.

I assumed while solving this that each side must've 4 people because otherwise there was no way to answer this question. But GMAT is likely to give you this info.

A boat has a crew of eight. Two of those eight can row only on the stroke side, while three can row only on the bow side. In how many ways can the two sides of the boat be manned if each side must have 4 people?

Then the question was whether '...boat be manned' requires us to give arrangements in which these people will sit. Since there was no option that gave answer as 3 (Select 2 of the 3 remaining people for stroke side in 3 ways), it meant that arrangements are required and that is how I reached 1728.

Given: A boat has a crew of eight. Two of those eight can row only on the stroke side, while three can row only on the bow side.

Asked: In how many ways can the two sides of the boat be manned?

The number of arrangements for stroke side crew = 4P2 = 4!/2! = 12
The number of arrangements for bow side crew = 4P3 = 4!/1! = 24
The number of arrangements for remaining 3 crew = 3! = 6

Total number of arrangement for 8 persons crew = 12*24*6 = 1728

IMO E

Alternate explanation
so we need 4 on bow side 4 on stroke side.

Let * people be 1,2,3,4,5,6,7,8
For bow side:

Let's say, 1,2,3 can be on bow side only so we have 1 2 3 __
For last blank we can choose from 3 ( 2 people who row on stroke side) = 3C1
now for bow side = selection*arrangement = 3C1*4!

For Stroke side:

Let's say, 4,5 can be on stroke side only so we have 4 5 __ __
so for two empty positions we have two people left/
hence, for stroke side = selction*arrangement = 2C2*4!

Answere therefore = 3C1*4!*2C2*4! = 1728 (E)

In my opinion, a very poorly worded question, because it requires specific knowledge about rowing.

It's not just the ambiguity in terminology ("bow" and "stroke" sides). The ambiguity is even more fundamental - how can we assume that there need to be 4 people on each side?

Another ambiguity is whether the bow and stroke sides are interchangeable or not. If yes, that would double the answer. But if they are fixed (with reference to something external, e.g., the "front" or "back" of the boat, or the "cox" as shown in Bunuel's illustration), then 1728 is fine.

An official GMAT question NEVER requires any such assumptions to be made.

I believe that these aspects should be made explicit, with no room for ambiguity, or this question should be taken off the forum.

To answer your first question, this question is treated as a permutation because PEOPLE are always distinct, different entities - they are NOT inter-changeable. One way to think about it is to give them names. For example, XY is counted as a different arrangement than YX.