menontion
chetan2u
Bunuel
A camp director must seat 50 campers in separate circles surrounding 3 different campfires. Twenty-five campers will sit around the first fire, 20 campers around the second fire, and 5 campers around the third fire. If the director must seat the campers around campfire 1 first, campfire 2 second, and campfire 3 third and if two seating arrangements are said to be different only when the positions of campers are different relative to each other, how many possible ways are there for him to seat the campers?
A. \((\frac{50!}{(25!25!)})(24!)(\frac{25!}{(5!20!)})\)
B. \((\frac{50!}{(25!25!)})(24!)(\frac{25!}{(5!20!)})(19!)(3!)\)
C. \((\frac{50!}{(25!25!)})(24!)(\frac{25!}{(5!20!)})(19!)(4!)\)
D. \((\frac{50!}{(25!25!)})(25!)(\frac{25!}{(5!20!)})(20!)(5!)\)
E. \((\frac{50!}{(25!25!)})(25!)(\frac{25!}{(5!20!)})(20!)(4!)\)
Remember these Q require more than one step..
1) first choose 25 out of 50 = 50C25..
--- These 25 can be placed in (25-1)! ways in a circle..2) second choose 20 out of 25 = 25C20
--- these 20 can be placed in (20-1)! ways in a circle3) third choose 5 out of 5 = 5C5
place these 5 in (5-1)! ways..total ways
50C25 * 24! * 25C20 * 19! * 5C5 * 4!..
\((\frac{50!}{(25!25!)})(24!)(\frac{25!}{(5!20!)})(19!)(4!)\)
ans C
Hey! I didn't get the (25-1), (20-1) and (5-1). Like logically, why are we doing this?
Thanks!
Hi..
This is a formula for combination in a circle..
In a straight line, if n people are arranged, ways are n!
But in a circle, it becomes (n-1)!..
Reason:-
In a circle, the arrangements are counted as SAME when we shift each by one chair.
Say there are three chairs I,II,III and three persons a, b and c.
In a straight line I-a, II-b and III-c and I-b, II-c, and III-a will be different....
But in circles, it will remain the same ... abc and bca and cab are all same. You cannot differentiate in all three unless the chairs are also different and different chairs mean different arrangement.
Hope it helps...