Question

in a boat there are 11 crew where 8 can bow but not steer 3 can steer but not bow suppose 2 of the bowers have to be on the same side

Answer 1

86400

Answer 2

The problem asks us to find the number of ways to arrange crew members in a boat given specific conditions.

Crew Breakdown:

• Total crew members: 11
• Bow (row) but not steer: 8 members (let's call this group B)
• Steer but not bow (row): 3 members (let's call this group S)

Boat Configuration: A standard rowing boat (like an eight-oared shell) typically has 8 rowers and 1 steerer (coxswain).

• The 8 members of group B are the available rowers. Since there are 8 rowing positions, all 8 B members will be used as rowers.
• The 3 members of group S are the available steerers. Only one steerer is needed for the boat. The other 2 S members will not be in the boat.

Therefore, the boat will consist of 1 steerer and 8 rowers, making a total of 9 crew members in the boat.

Step 1: Select the Steerer One steerer must be chosen from the 3 members of group S. Number of ways to select the steerer = $^3C_1 = 3$ ways.

Step 2: Arrange the Rowers There are 8 rowers (all members of group B). These 8 rowers need to be arranged in 8 distinct rowing positions. These positions are typically split equally on two sides of the boat (e.g., 4 on the port side, 4 on the starboard side). The positions within each side are also distinct (e.g., stroke, 7, 6, 5, 4, 3, 2, bow).

The condition is: "2 of the bowers have to be on the same side". This implies that a specific pair of bowers (let's call them $B_1$ and $B_2$) must be on the same side. If it meant "at least one pair of bowers", the condition would be trivially true for 4 rowers on each side.

Let's calculate the number of ways to arrange the 8 distinct rowers in the 8 distinct positions such that two specific rowers ($B_1$ and $B_2$) are on the same side. We can use the principle of complementary counting: Total arrangements = (Total arrangements without any restriction) - (Arrangements where $B_1$ and $B_2$ are on opposite sides).

• Total arrangements of 8 distinct rowers in 8 distinct positions: This is $8! = 40,320$ ways.

• Arrangements where $B_1$ and $B_2$ are on opposite sides:

  1. Choose a side for $B_1$ (2 options: Port or Starboard). Let's say $B_1$ is on Port.
  2. Choose a position for $B_1$ on that side (4 positions available).
  3. $B_2$ must be on the opposite side. Choose a position for $B_2$ on the opposite side (4 positions available).
  4. Arrange the remaining 6 rowers in the remaining 6 positions. This can be done in $6!$ ways.

  So, the number of ways $B_1$ and $B_2$ are on opposite sides is: $(4 \text{ pos for } B_1 \text{ on Side A}) \times (4 \text{ pos for } B_2 \text{ on Side B}) \times (6! \text{ for remaining}) = 4 \times 4 \times 6! = 16 \times 720 = 11,520$ ways. (Note: We already accounted for $B_1$ on Port and $B_2$ on Starboard. If we considered $B_1$ on Starboard and $B_2$ on Port, it would be another $4 \times 4 \times 6!$. So, $2 \times 4 \times 4 \times 6!$ if we choose sides first, then positions. However, the positions are distinct, so choosing a position for $B_1$ on one side and a position for $B_2$ on the other side directly accounts for this. If $B_1$ is in Port-1 and $B_2$ in Starboard-1, that's one arrangement. If $B_1$ is in Starboard-1 and $B_2$ in Port-1, that's another. The $4 \times 4$ already accounts for which of the 4 positions on each side they occupy. The $8!$ total arrangements already distinguishes all positions. So, the $4 \times 4 \times 6!$ is correct for a specific pair being on opposite sides.)

• Arrangements where $B_1$ and $B_2$ are on the same side: Total arrangements - Arrangements where $B_1$ and $B_2$ are on opposite sides $= 8! - (4 \times 4 \times 6!) = 40,320 - 11,520 = 28,800$ ways.

Step 3: Combine the selections and arrangements Total number of ways = (Ways to select steerer) $\times$ (Ways to arrange rowers under the given condition) Total ways = $3 \times 28,800 = 86,400$.

The question is incomplete as it doesn't explicitly ask "In how many ways can they be arranged?". Based on the context of similar problems, it is implied.