Question

A boat is to be manned by eight men of whom 2 can only row on bow side and 3 can only row on stroke side, the number of ways in which the crew can be arranged is
A. 4360
B. 5760
C. 5930
D. None of these

Answer 1

Hint: The eight men are splitted into 4 men on the bow side and 4 men on the stroke side.
We use permutation and combination formulas to find the number of ways.
First we use a combination formula to find the number of ways in which the 2 men can be arranged on the bow side.
Also, we use a combination formula to find the number of ways in which the 3 men can be arranged on the stroke side.

Formula Used: \[{}^n\mathop P\nolimits_n = n!\] , \[{}^n\mathop C\nolimits_r = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]
Here n is the total number of objects and r is the number of objects selected.

Complete step by step solution: Given that the boat is manned by eight men.
We should place four men on the bow side and another four men on the stroke side.
Also, given that only two men can row on the bow side and only three can row on the stroke side.
Calculating the number of ways of arranging only two men on bow side and three men on stroke side:
We have to select only two men on the bow side and only three men on the stroke side from the 5 men.
Number of ways \[ = {}^5\mathop C\nolimits_2 \times {}^3\mathop C\nolimits_3 \]
Calculating the number of ways of arranging 4 men on both the sides of the boat:
Number of ways of arranging 4 men on bow side \[ = {}^4\mathop P\nolimits_4 \]
Number of ways of arranging 4 men on stroke side \[ = {}^4\mathop P\nolimits_4 \]
Calculating the total number of ways:
Total number of ways \[ = {}^5\mathop C\nolimits_2 \times {}^3\mathop C\nolimits_3 \times {}^4\mathop P\nolimits_4 \times {}^4\mathop P\nolimits_4 \]
\[{}^5\mathop C\nolimits_2 \times {}^3\mathop C\nolimits_3 \times {}^4\mathop P\nolimits_4 \times {}^4\mathop P\nolimits_4 = \dfrac{{5!}}{{2!\left( {5 - 2} \right)!}} \times \dfrac{{3!}}{{3!\left( {3 - 3} \right)!}} \times 4! \times 4!\]
 \[\begin{array}{l} = \dfrac{{5!}}{{2!3!}} \times \dfrac{{3!}}{{3!0!}} \times 4! \times 4!\\ = \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{\left( {2 \times 1} \right)\left( {3 \times 2 \times 1} \right)}} \times \dfrac{{3!}}{{3!\left( 1 \right)}} \times \left( {4 \times 3 \times 2 \times 1} \right) \times \left( {4 \times 3 \times 2 \times 1} \right)\\ = 10 \times 1 \times 24 \times 24\\ = 5760\end{array}\]
Therefore, the number of ways is 5760.


Option ‘B’ is correct

Note: First we will have to find the number of ways of arranging only two men on bow side and only three on stoke side.
The arrangement of four on each side is calculated using permutation.