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17 Feb 2021, 09:00
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B
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Difficulty:
95%
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Question Stats:
34% (02:11) correct
66%
(02:13)
wrong
based on 50
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There are counters available in 3 different colors (at least four of each color). Counters are all alike except for the color. If ‘m’ denotes the number of arrangements of four counters if no arrangement consists of counters of same color and ‘n’ denotes the corresponding figure when every arrangement consists of counters of each color, then:
(A) m = 2n
(B) 6m = 13n
(C) 3m = 5n
(D) 5m = 3n
(E) 5m = 7n
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
(A) m = 2n
(B) 6m = 13n
(C) 3m = 5n
(D) 5m = 3n
(E) 5m = 7n
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
17 Feb 2021, 17:11
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Total possible arrangements of counters = 3^4=81
Case 1- Number of cases when the all counters consist of same color= 3
m= 81-3 = 78
Case 2- when every arrangement consists of counters of each color
If this case 2 counters have same color and other 2 have different one
Total possible arrangements, n = 3C1* 4!/2! = 36
B
Case 1- Number of cases when the all counters consist of same color= 3
m= 81-3 = 78
Case 2- when every arrangement consists of counters of each color
If this case 2 counters have same color and other 2 have different one
Total possible arrangements, n = 3C1* 4!/2! = 36
B
Bunuel
07 Mar 2021, 07:22
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Bunuel
Solution:
We see that there are 3 colors and there are at least 4 counters of each color. Let’s assume there are exactly 4 of each color. The first counter can be any of 3 colors, the second counter can be any of 3 colors, the third counter can be any of 3 colors, and the fourth counter can be any of 3 colors. Thus, there is a total of 3 x 3 x 3 x 3 = 3^4 = 81 different arrangements of the counters.
We are told that m = the number of arrangements of the counters such that the arrangement isn’t all the same color. Because there are 3 colors (assume red, blue, and green), we subtract out the 3 occurrences where the 4 counters are all the same color. (all red or all blue or all green, for a total of 3 arrangements). Thus, m = 81 - 3 = 78,
We are told that n = the number of arrangements in which each of the 3 colors must be in the arrangement. This means that an acceptable arrangement must include 1 red, 1 blue, and 1 green, allowing for the fourth counter to be any color.
Let’s first find the number of arrangements containing 2 red counters, 1 blue counter and 1 green counter. There are 4C2 = 4!/(2!*2!) = 6 choices for the red counter. Since there are two remaining counters, there are 2 choices for the blue counter. The remaining counter will be the green counter. Thus, there are 6 x 2 = 12 arrangements that contain 2 red counters, 1 blue counter and 1 green counter.
Next, notice that in addition to the scenario we considered above, we can also have 1 red counter, 2 blue counters and 1 green counter, or 1 red counter, 1 blue counter and 2 green counters. A similar calculation can be applied to each of these scenarios, so they all contain 12 arrangements. Thus, in total, there are 3 x 12 = 36 arrangements containing each of the three colors. In other words, n = 36.
Of the given relations, only 6m = 13n is satisfied when m = 78 and n = 36.
Answer: B
