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18 May 2019, 13:51
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
53% (01:39) correct
47%
(01:40)
wrong
based on 378
sessions
History
Date
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In how many ways can one divide twelve different chocolate bars equally among four boys?
A) 12!/3!
B) 12!/4!
C) 12!/(3!)^4
D) 12!/(4!)^3
E) 12!/(3!4!)
A) 12!/3!
B) 12!/4!
C) 12!/(3!)^4
D) 12!/(4!)^3
E) 12!/(3!4!)
dabaobao
TL;DR
Using Anagram Grid:
AAA BBB CCC DDD
12!/(3!3!3!3!) = 12!/(3!)^4
Official Solution
Credit: Veritas Prep
You need to divide 12 chocolate bars among four boys i.e. you have to make four distinct groups. To boy 1, you can give the first chocolate in 12 ways, the second chocolate in 11 ways and the third chocolate in 10 ways. But we don’t want to arrange the chocolates so you can select 3 chocolates for boy 1 in 12*11*10/3! ways (this is equivalent to 12C3 if you follow the formula). Similarly, you can select 3 chocolates for the second boy in 9*8*7/3! ways (i.e. 9C3), for the third boy in 6*5*4/3! ways (i.e. 6C3) and for the fourth boy in 3*2*1/3! ways (i.e. 3C3)
Therefore, you can distribute 12 chocolates among 4 boys equally in (12*11*10/3!) * (9*8*7/3!) * (6*5*4/3!) * (3*2*1/3!) = 12!/(3!*3!*3!*3!) ways
Alternatively, you can visualize putting the 12 chocolates in a row in 12! ways and drawing a line after every three chocolates to demarcate the groups.
OOO l OOO l OOO l OOO
Since the chocolates within the groups are not arranged, we divide 12! by 3! for every group. Since there are 4 groups, number of ways of making 4 distinct groups = 12!/(3!*3!*3!*3!)
ANSWER: C
General Discussion
21 May 2019, 06:13
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dabaobao why don't we need to divide the whole thing by 4!?
I was thinking the solution should be 12!/[(3!^4)*4!]
I was thinking the solution should be 12!/[(3!^4)*4!]
