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  • Qu124 - Boating Party
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Posted October 23, 2015 by in Bonus Probability Questions 2351

Qu124 - Boating Party

“What a lovely morning for a boating party,” said Professor Sinker, as the party of 12 made their way down to the mooring, “the light is Caravaggesque.” The light was indeed spectacular, the dark clouds of a gathering storm contrasting splendidly with morning light, and with steep waves on the lake adding a little drama to the scene. They walked on for a moment, then Professor Downer, who was clearly troubled by the Caravaggio reference, turned to Sinker and said “I think you mean Géricault. It's the spitting image of The Raft of the Medusa.” They continued towards the lake.

“I believe I saw that in the Louvre,” said Colonel Cadaver. “Is it the one where they ate each other?”

“The very same,” said Downer. “clearly the subulty of the work was not lost on you.”

At that moment they arrived at the mooring to find four boats, each of which could take three people. “Only four of us including myself know how to row,” said Sinker, “and I always take this boat.” He pointed to the only boat in a sound state of repair. “No need to worry though,” he continued, “once we bail these other three boats out they will be perfectly servicable. We just need to work out who is going in which boat.”

How many ways are there of arranging a boating party of 12 people into four different boats containing three people each if only four of the party are oarsmen, each boat must contain an oarsman, and if one of the oarsmen must be in a particular boat?

Qu124 - Image 01pngAnswer:

Let us first consider how many ways the four oarsmen can be distributed amongst the boats. Professor Sinker must be in a particular boat so his position is uniquely determined. For the second boat we have a choice of 3 oarsmen, for the third boat 2 oarsmen, and thus for the fourth boat we have no choice at all and it must be occupied by the remaining oarsman. Thus the four oarsmen can occupy the boats in 1 × 3 × 2 × 1=3!=6  ways.
Now we have four boats each with two empty seats, and 8 remaining members of the boating party to place in those seats. Consider the two seats in the first boat. For the first seat we have a choice of 8 people, and for the second seat 7 people. Thus there are 8 × 7 ways of filling those seats. We must not forget, however, that the order in which we fill the seats does not affect how the boat is filled—that is, the pair that fill the seats could swap seats and the composition of the boat would remain the same. The number of unique ways of filling the boat becomes ( 8 × 7 )/2!
The ways of filling the second, third and fourth boats follow by analogy. They are ( 6 × 5 )/2! , ( 4 × 3 )/2! , and ( 2 × 1 )/2! . Thus the total number of ways of arranging the boating party is: n=3! (( 8 × 7 )/ 2!) (( 6 × 5 )/ 2!) (( 4 × 3 )/ 2!) (( 2 × 1 )/ 2!) = (3!)(8!) /(2!)^4 =15,120

 

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