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25 Mar 2016, 12:39
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
16% (01:41) correct
84%
(02:44)
wrong
based on 31
sessions
History
Date
Time
Result
Not Attempted Yet
Five kids were walking down the street. At a shop, they saw three brands of chocolates viz. Cadbury, Nestle and Ferrero being displayed. The shopkeeper has three chocolates which are of three different flavors of each brand. The kids bought all the chocolates available and decided to play a game. The chocolates will be distributed only to the winners of that game. Also they have decided that any winner cannot have all three brands of chocolates as his prize. Find the number of ways in which the chocolates can be distributed when there are four winners of the game and the best of them gets at least 4 chocolates?
A.12312
B.22536
C.4320
D.11268
E.45072
OE will be posted after 1 week on 1st april 2016
A.12312
B.22536
C.4320
D.11268
E.45072
OE will be posted after 1 week on 1st april 2016
26 Mar 2016, 01:04
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rhine29388
Hi,
some points on the Q..
1) firstly, it can never be a GMAT Q. You do not expect such complicated and cumbersome calculations in ACTUALS. Also, the wordings are much clearer and crisp in ACTUALS.
2) The "the best of them gets at least 4 chocolates" in actuals would mean that there is one specific kid who is the best of them. But when you solve it, it can be any of the four and then only you will get the OA.
3) Although it is not a GMAT Q, and you will be wasting time on it as I wasted 4 to 5 minutes on it due to ambiguous details and then more number of steps, the solution will be found in different cases
SOLUTION
CASE I- FOUR to the winner
4 - 3-1-1
a) four can be taken in 3 of one kind and 1 of other kind and these two can be choosen out of three brands in 3C2= 3..
3 for the second can be 2 of one brand and 1 of other OR all three of one brand
4-2-2-1
a) same case for 4 choclates
b) 2 can be from one brand OR from two different brands
CASE II- FIVE to the winner
5-2-1-1
a) again these 5 can be any one of the four and can be from any two brands-4C1*3C2*3!=36
b) 2 for the second can be from same brand or two different brands=3*3!=18
c) 1 each can be in two ways
total= 36*18*2= 1296 ways
CASE III- 6 to the winner
6-1-1-1
a) again these 6 can be given to any four and from any two brands
4C1*3C2*3!=4*3*3*2=72 ways
For knowledge, one can see the solution and learn, BUT for GMAT just SKIP it
31 Mar 2016, 13:16
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Official Solution
Since, one child cannot have more than two brands of chocolates, therefore, one child can get a maximum of six chocolates.
When there are four winners, the winners can have the chocolates in the following possible way
Case I
One of them gets six chocolates and the other three get one each.
The one who gets six chocolates can be selected in 4C1 ways and he will get any two brands out of three in 3C2 ways.
The three chocolates of the same brand will be distributed among the other three in 3! ways.
Hence, the total number of ways = 4 × 3 × 6 = 72
Case II
One of them gets 5 chocolates, one of them gets 2 chocolates and the other two get 1 each.
Kid with 5 chocolates:
The one who gets five chocolates can be selected in 4C1 ways and he will get any two brands out of three in 3C2 ways.
After choosing the two brands (say A & B), he can get all three chocolates of one brand in 2 ways (i.e. either A or B). The other two chocolates can be chosen from the second brand in 3 ways. Thus, there are 4 × 3 × 2 × 3 = 36 ways
Kid with 2 chocolates:
This person can be chosen in 3 ways. Now, he can either get two chocolates from the same brand or from two different brands.
Thus, there are 6 × 3 = 18 ways
Kids with 1 chocolate each
The remaining chocolates can be distributed between the two remaining winners in 2 ways.
Therefore, total number of ways = 36 × 18 × 2 = 1296 ways.
Case III
Similarly for four chocolates the following combinations are possible:
a) 4 3 1 1
The number of ways = 4 × 3 × [{6 × 3(1 + ( 6 + 3))} + {1 × 9 × 3 (1 + 6)}] × 2 = 8856
b) 4 2 2 1
The number of ways = 4 × 3 × [{6 × 3(2 × 6 × 3 + (3 + 9))} + {9 × (9 + 9)}] = 12312
Since, the best of them can’t have less than 4 chocolates the above are the only cases applicable.
Hence, the total number of ways = 72 + 1296 + (8856 + 12312) = 22536.
Since, one child cannot have more than two brands of chocolates, therefore, one child can get a maximum of six chocolates.
When there are four winners, the winners can have the chocolates in the following possible way
Case I
One of them gets six chocolates and the other three get one each.
The one who gets six chocolates can be selected in 4C1 ways and he will get any two brands out of three in 3C2 ways.
The three chocolates of the same brand will be distributed among the other three in 3! ways.
Hence, the total number of ways = 4 × 3 × 6 = 72
Case II
One of them gets 5 chocolates, one of them gets 2 chocolates and the other two get 1 each.
Kid with 5 chocolates:
The one who gets five chocolates can be selected in 4C1 ways and he will get any two brands out of three in 3C2 ways.
After choosing the two brands (say A & B), he can get all three chocolates of one brand in 2 ways (i.e. either A or B). The other two chocolates can be chosen from the second brand in 3 ways. Thus, there are 4 × 3 × 2 × 3 = 36 ways
Kid with 2 chocolates:
This person can be chosen in 3 ways. Now, he can either get two chocolates from the same brand or from two different brands.
Thus, there are 6 × 3 = 18 ways
Kids with 1 chocolate each
The remaining chocolates can be distributed between the two remaining winners in 2 ways.
Therefore, total number of ways = 36 × 18 × 2 = 1296 ways.
Case III
Similarly for four chocolates the following combinations are possible:
a) 4 3 1 1
The number of ways = 4 × 3 × [{6 × 3(1 + ( 6 + 3))} + {1 × 9 × 3 (1 + 6)}] × 2 = 8856
b) 4 2 2 1
The number of ways = 4 × 3 × [{6 × 3(2 × 6 × 3 + (3 + 9))} + {9 × (9 + 9)}] = 12312
Since, the best of them can’t have less than 4 chocolates the above are the only cases applicable.
Hence, the total number of ways = 72 + 1296 + (8856 + 12312) = 22536.
