There are counters available in x different colors. The counters are

Question

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

(A) x^y – x 
(B) x^y – y
(C) y^x – x 
(D) y^x – y
(E) y^x – 2y

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ravigupta2912 · Answer

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.

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nikospanayo · Answer

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

ravigupta2912 · Answer

You’re right. My answer stands modified. I’ll consider my original post as an error.

ScottTargetTestPrep · Answer

Solution:

The first counter can be painted with x possible colors. The second counter can be painted with x possible colors. If there were just 2 counters, we would have x^2 different arrangements. If there were 3 counters, we would have x^3 different arrangements.

Since we have y counters, we see that the number of arrangements of counters is x^y, including those that have all counters the same color. However, if the counters can’t be all the same color, we need to subtract x arrangements that have all the counters the same color. That is, there are x^y - x arrangements of counters where not all of them have the same color.

Answer: A

Fdambro294 · Answer

Total number of arrangements in which the color of the counter can be repeated

=

(no. of arrangements in which counters are NOT all the same color)

+

(no. of arrangements in which counters ARE all the same color)

If there are X different color counters and Y “slots” in which the colors can be arranged, then first (assuming colors can repeat themselves)

Slot 1: X options

Slot 2: X options

Slot 3: X options

.....

All the way to Slot Y: X options

(X) multiplied by itself Y times will give us every possible arrangement in which the colors CAN be repeated ——- (X)^y

From this, we want to remove the Unfavorable Outcomes. These include the arrangements in which every “slot” has the same color counter.

If there is X different colors, the number of these arrangements will be:

1st color repeated for Y “slots”

2nd color repeated for Y “slots

....

All the way up to the Xth color repeated for Y “slots”

This means we will have exactly X arrangements in which every counter picked will be the same color.

Answer:

(X)^y - (X)

Answer A

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bumpbot · Answer

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There are counters available in x different colors. The counters are

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There are counters available in x different colors. The counters are

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