Question

Out of a crew of 43 persons on a ship, only 3 can do the steering and one is needed. Of the remaining 40, 8 persons wish to go only on one side and 3 persons wish to go on the other side. In how many ways can they be arranged so that there are 20 rowers on each side of the ship?

Answer 1

Hint: First select 1 person from the three persons who can steer the ship using the formula of combinations ${}^{3}{{C}_{1}}$. Now, consider that row 1 has already 8 rowers and row 2 has 3 rowers. Select the remaining 12 rowers from 29 rowers left needed for row 1 by using the relation ${}^{29}{{C}_{12}}$. The remaining 17 will automatically get selected for row 2. Finally, arrange the 20 rowers among themselves present in each row by using the relation $20!\times 20!$. Multiply all the expressions to get the answer.

Complete step by step answer:
First of all we need to select 1 person who will steer the ship out of three persons who can do the steering. So we have,
$\Rightarrow $ Number of ways to select 1 person from 3 persons for steering = ${}^{3}{{C}_{1}}$
Now, it is said that there should be 20 rowers on the sides of the boat. Also, 8 rowers have their own wish to go to one side and 3 rowers have their wish to go on the other. Let us assume that 8 rowers are in row 1 and 3 are in row 2. So, 29 rowers are left in which row 1 must have 12 more rowers and the remaining 17 must be in row 2.
$\Rightarrow $ Number of ways to select 12 persons from 29 persons for row 1 = ${}^{29}{{C}_{12}}$
Therefore the remaining 17 will automatically get selected for row 2.
So we have selected the rowers for both sides, now we need to arrange them. So we get,
$\Rightarrow $ Number of ways in which 20 rowers can be arranged among themselves in row 1 = $20!$
$\Rightarrow $ Number of ways in which 20 rowers can be arranged among themselves in row 2 = $20!$
Therefore, the total number of arrangements that can be made = ${}^{3}{{C}_{1}}\times {}^{29}{{C}_{12}}\times 20!\times 20!$

Note: Note that we haven’t applied the combination formula for the selection of 17 people that is needed in row 2 because we have already selected 12 rowers out of 29 for row 1 and therefore the remaining rowers are 17 in which we have to select 17 so there will be only one way of selection given as $^{17}{{C}_{17}}$. The mathematical reason for such a condition is due to the formula $^{n}{{C}_{r}}{{=}^{n}}{{C}_{n-r}}$.