The game of chess involves 16 pieces for each player: 1 king, 1 queen, 2 rooks, 2 knights, 2 bishops, and 8 pawns.

Assuming SIVA is placing his pieces on the first two rows of his side of the board, the probability of him placing all pieces in the correct position can be calculated as follows:

1. There is only 1 correct place for the king and 1 correct place for the queen.
2. There are 2 correct places for the rooks, 2 for the knights, and 2 for the bishops. However, these pieces are indistinguishable from each other (i.e., both rooks are the same, both knights are the same, etc.). Therefore, the number of ways to arrange these pieces is 2! (factorial) for each type of piece.
3. There are 8 correct places for the pawns, but again, these pieces are indistinguishable from each other. Therefore, the number of ways to arrange these pieces is 8!.

The total number of ways to arrange all the pieces is therefore (1*1)*(2!*2!*2!)*8! = 3,715,891,200.

However, there is only 1 way to arrange the pieces correctly. Therefore, the probability of SIVA placing all pieces in the correct position is 1/3,715,891,200, which is virtually zero.

Question

The game of chess involves 16 pieces for each player: 1 king, 1 queen, 2 rooks, 2 knights, 2 bishops, and 8 pawns.

Assuming SIVA is placing his pieces on the first two rows of his side of the board, the probability of him placing all pieces in the correct position can be calculated as follows:

1. There is only 1 correct place for the king and 1 correct place for the queen. 
2. There are 2 correct places for the rooks, 2 for the knights, and 2 for the bishops. However, these pieces are indistinguishable from each other (i.e., both rooks are the same, both knights are the same, etc.). Therefore, the number of ways to arrange these pieces is 2! (factorial) for each type of piece.
3. There are 8 correct places for the pawns, but again, these pieces are indistinguishable from each other. Therefore, the number of ways to arrange these pieces is 8!.

The total number of ways to arrange all the pieces is therefore (1*1)*(2!*2!*2!)*8! = 3,715,891,200.

However, there is only 1 way to arrange the pieces correctly. Therefore, the probability of SIVA placing all pieces in the correct position is 1/3,715,891,200, which is virtually zero.

Knowee AI · Accepted Answer

The game of chess involves 16 pieces for each player: 1 king, 1 queen, 2 rooks, 2 knights, 2 bishops, and 8 pawns.

Assuming SIVA is placing his pieces on the first two rows of his side of the board, the probability of him placing all pieces in the correct position can be calculated as follows:

1. There is only 1 correct place for the king and 1 correct place for the queen. 
2. There are 2 correct places for the rooks, 2 for the knights, and 2 for the bishops. However, these pieces are indistinguishable from each other (i.e., both rooks are the same, both knights are the same, etc.). Therefore, the number of ways to arrange these pieces is 2! (factorial) for each type of piece.
3. There are 8 correct places for the pawns, but again, these pieces are indistinguishable from each other. Therefore, the number of ways to arrange these pieces is 8!.

The total number of ways to arrange all the pieces is therefore (1*1)*(2!*2!*2!)*8! = 3,715,891,200.

However, there is only 1 way to arrange the pieces correctly. Therefore, the probability of SIVA placing all pieces in the correct position is 1/3,715,891,200, which is virtually zero.

SIVA and Shyam are playing chess together. SIVA known the two rows in which he has to put all the pieces in but he doesn’t know how to place them. What is the probability that he puts all the pieces in the right place?

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