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17 Mar 2015, 05:45
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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%
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Question Stats:
28% (02:46) correct
72%
(02:39)
wrong
based on 166
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Suppose we have six marbles: 3 blue marbles, 2 red marbles, and one green marble. Suppose we are going to put them into three cups: a black cup, a white cup, and a purple cup. We could put all six in any cup and leave two cups empty; or we could put marbles in two cups and leave one cup empty; or we could put some marbles in each of the three cups. How many combinations are possible?
(A) 90
(B) 180
(C) 360
(D) 540
(E) 720
(A) 90
(B) 180
(C) 360
(D) 540
(E) 720
23 Mar 2015, 04:01
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Bunuel
Suppose we have six marbles: 3 blue marbles, 2 red marbles, and one green marble. Suppose we are going to put them into three cups: a black cup, a white cup, and a purple cup. We could put all six in any cup and leave two cups empty; or we could put marbles in two cups and leave one cup empty; or we could put some marbles in each of the three cups. How many combinations are possible?
(A) 90
(B) 180
(C) 360
(D) 540
(E) 720
Kudos for a correct solution.
(A) 90
(B) 180
(C) 360
(D) 540
(E) 720
Kudos for a correct solution.
MAGOOSH OFFICIAL SOLUTION:
First, consider possibilities the 3 blue marbles.
Blue Case I: all three in one cup = 3 possibilities
Blue Case II: two in one cup, one in another è three choices for the cup with two, then two choices for the cup with one = 6 possibilities
Blue Case III: one in each cup = 1 possibility
total = 3 + 6 + 1 = 10 possibilities for the blue marbles
Now, the 2 red marbles.
Red Case I: two in one cup = 3 possibilities
Red Case II: one in cup, one in another = 3 possibilities for which cup is empty
total = 6 possibilities for the red marbles
Now, the green marble can simply go in one of the three cups = 3 possibilities.
By the Fundamental Counting Principle, multiple the blue & red & green possibilities:
N = 10*6*3 = 180
Answer = (B)
17 Mar 2015, 09:37
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Bunuel
Suppose we have six marbles: 3 blue marbles, 2 red marbles, and one green marble. Suppose we are going to put them into three cups: a black cup, a white cup, and a purple cup. We could put all six in any cup and leave two cups empty; or we could put marbles in two cups and leave one cup empty; or we could put some marbles in each of the three cups. How many combinations are possible?
(A) 90
(B) 180
(C) 360
(D) 540
(E) 720
Kudos for a correct solution.
(A) 90
(B) 180
(C) 360
(D) 540
(E) 720
Kudos for a correct solution.
# ways to distribute Blue marbels(3) is :
1 in each cup= 1
2 in one cup and 1 in other=6
all 3 in one cup = 3
total 1+6+3=10 possibilities
# ways to distribute Red marbels(2) is :
1 in each cup = 3 ways
all in one cup = 3ways
total 6ways.
# ways to distribute Green marbels(1) is :
3 ways
total number of ways = 10*6 *3 = 180
Alternately this can be solved using combitronics with possible distribution
(6,0,0) one cup receives all = 3ways
(5,1,0)= one cup receives 5 and other receives 1 ; 3C1 * 2 but we will have to multiply this by 3! ; total = 3*2*6=36
(4,1,1)= one cup receives 4 and other receive 1 each ; 4C1 *2*1 but we will have to multiply this by 3!/2! ; total =24
like wise if we do for
4,2,0
3,3,0
3,2,1
2,2,2
we will get same total 180 ways.
Answer is B.
