A boat is to be manned by 10 men of whom, one cannot row on the bow si

The "bow" of a boat is the front of the boat. So it actually doesn't make any sense to assign half your rowers to the "bow side" of the boat (where are the other half, at the stern? In what direction is the boat going?). Rowers are normally on the left and right side of the boat, not at the front and the back.

Regardless, the question absolutely needs to tell you how you're allowed to divide the rowers between the two sides. It also needs to tell you if order matters - are we just assigning our rowers to the two sides of the boat, or are we arranging them from front to back as well? I wouldn't suggest studying questions written like this; no real GMAT question would ever be written this way, and if you need to guess what the question writer means, it's going to be mostly luck whether you're even answering the question they intended to ask. Where is this from?

Here we have 1 person who must be assigned to the non-bow side, and 3 people who must be assigned to the bow side. That leaves us only 6 people we can freely assign to either side. Assuming we need 5 on each side, we have 6C2 ways to pick the final two people for the bow side, or (6)(5)/2! = 15 possible selections. The remaining people then all go to the non-bow side. Once we've made those selections, we'll be able to put the bow side rowers in 5! possible orders, and same with the rowers on the other side, so the answer is (15)(5!)(5!) = 216,000. But I'm making so many guesses about what the question writer meant that I wouldn't be at all surprised to learn that the "OA" is something else.

Normally, when a boat is rowed, you would have equal men on both sides.

Rakshit TN, If this piece of data is added, then the question would be complete (based on the OA given)

If A be the person who cannot be on the bow side, so he will be on the opposite side. Let B, C and D be the three oarsman who have to be on the bow side (refer to the diagram below)

There are 6 oarsmen remaining who have no restrictions. We need 2 out of these on the bow and these can be chosen in 6C2 = 15 ways

The remaining 4 will go to the other side in 4C4 = 1 way.

On each side the 5 oarsmen can be arranged in 5! = 120 ways

Therefore total number of ways = 15 * 1 = 120 * 120 = 216000


Option E

Arun Kumar

Yes, modified. Thanks

The question is from the Practice set from TIME institute. Question thread is modified now. Thanks

Odd wording aside, another one that can be solved with multiplying out all the options where the men can sit, starting with the men who have restrictions.

The first man only has 5 options to choose from on the "not bow" side.
The second has 5 options to choose from on the "bow" side
The third has 4 options (since the second has taken one of the options)
The fourth has 3 options (since the second and third have taken an option)
Now the rest can sit anywhere else on the boat, which has 6 seats left. So 6 options, 5 options, 4 options, etc as the men take up seats.

Therefore, multiply the above:
5 * 5 * 4 * 3 * 6! = 216,000 (Answer E)

Given: A boat is to be manned by 10 men of whom, one cannot row on the bow side and three of the others can row only on the bow side.
Asked: In how many ways can the crew be arranged? (5 men can row on each side)

The number of ways the crew be selected in bow side = (10-1-3)C(5-3) = 6C2 = 15
Others will be used in non-bow side

The number of ways the crew be arranged = 15*5!5! = 15*120*120 = 216000

IMO E