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15 Feb 2021, 05:00
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
44% (02:12) correct
56%
(02:04)
wrong
based on 126
sessions
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There are counters available in x different colors. The counters are all alike except for the color. The total number of arrangements consisting of y counters, assuming sufficient number of counters of each color, if no arrangement consists of all counters of the same color is:
(A) x^y – x
(B) x^y – y
(C) y^x – x
(D) y^x – y
(E) y^x – 2y
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(A) x^y – x
(B) x^y – y
(C) y^x – x
(D) y^x – y
(E) y^x – 2y
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
ravigupta2912
15 Feb 2021, 09:49
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Independent events.
Each counter can be painted in x ways. Total number of counters is y
Total arrangements = y^x
Arrangements where each counter has the same colour = x
To visualise this better, i ascribed numbers 3 (x) colours for 5 (y) counters. There are only 3 possible arrangements where all the counters will have the same colour.
Therefore correct answer is y^x - x.
IMO C. Let’s see the OA.
Posted from my mobile device
Each counter can be painted in x ways. Total number of counters is y
Total arrangements = y^x
Arrangements where each counter has the same colour = x
To visualise this better, i ascribed numbers 3 (x) colours for 5 (y) counters. There are only 3 possible arrangements where all the counters will have the same colour.
Therefore correct answer is y^x - x.
IMO C. Let’s see the OA.
Posted from my mobile device
15 Feb 2021, 10:50
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ravigupta2912
i think total araangments are x^y we have y slots and each can have x different choices that gives x^y not y^x. i agrre with minus x
answer A
