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In how many ways can 4 white and 3 black chess pieces be

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by [email protected] » Tue Apr 17, 2018 7:03 pm
Hi All,

We're asked for the number of ways that 4 distinct white chess pieces and 3 distinct black chess pieces can be arranged in a row such that they occupy alternate places. This question is essentially one big Permutation question and can be solved with that type of calculation.

Since the pieces have to be in ALTERNATING colors, the arrangement must be WBWBWBW

For the 1st spot, we have 4 white pieces to choose from. Once we place one...
For the 2nd spot, we have 3 black pieces to choose from. Once we place one...
For the 3rd spot, we have 3 white pieces to choose from. Once we place one...
For the 4th spot, we have 2 black pieces to choose from. Once we place one...
For the 5th spot, we have 2 white pieces to choose from. Once we place one...
For the 6th spot, we have just 1 black piece to choose from.
For the 7th spot, we have just 1 white piece to choose from.

(4)(3)(3)(2)(2)(1)(1) = 144 possible arrangements.

Final Answer: B

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by Jeff@TargetTestPrep » Thu Apr 19, 2018 4:39 pm
BTGmoderatorLU wrote:In how many ways can 4 white and 3 black chess pieces be arranged in a row such that they occupy alternate places? Assume that the pieces are distinct.

A. 288
B. 144
C. 12
D. 48
E. 96
Since the chess pieces must occupy alternate places, the first piece must be white. In that case, the number of ways they can be arranged is :

4 x 3 x 3 x 2 x 2 x 1 x 1 = 144

Answer: B

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by deloitte247 » Sun Apr 22, 2018 10:03 am
Since the pieces have to be in alternating colors of the class piece, the arrangement = W B W B W B W ( where W= White piece and B= black piece)
4 white + 3 black piece= 7 (it will be on 7 spot)
(W) 1st spot= we can choose from 4 white pieces
(B) 2nd spot= we can choose from 3 black pieces
(W) 3rd spot= we can choose from 3 white pieces
(B) 4th spot = we can choose from 2 black pieces
(W) 5th spot= we can choose from 2 black pieces
(B)6th spot= we can choose from 1 black pieces
(W)7th spot=we can choose from 1 black pieces

Arrangement= (4 (3 (3)(2)(2)(1)(1)
= 4*3*3*2*2*1*1
=144 possible arrangements
Hence the answer is B
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