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# two pawns

Best Answer m00li, 21 April 2014 - 03:58 AM

there are 8 pawns, 2 rooks, 2 knights, 2 bishops, 1 king and 1 queen in white. similarly we have 8,2,2,2,1,1 sets of unique pieces in black.

Method 1:

A = number of ways of selecting 5 out of the above

B = number of ways of selecting 5 such that no pawn is selected

C = number of ways of selecting 5 such that only 1 or 0 white pawn is selected

A = coeff of x5 in (1-x9)2(1-x3)6(1+x)4(1-x)-8 = 2540

B = coeff of x5 in (1-x3)6(1+x)4(1-x)-6 = 876

C = coeff of x5 in (1+x)(1-x9)(1-x3)6(1+x)4(1-x)-7=2230

Method 2

A = no. of ways of selecting 5 such that 2 or 3 or 4 or 5 white pawns are selected

B = no. of ways of selecting 5

C = no. of ways of selecting 5 without any pawns

B = (all 5 pieces are similar) + (4 similar and 1 different) + (3 sim, 2 sim) + (3 sim,2 different) + (2 sim, 2 sim, 1 diff) + (2 sim, 3 diff) + (5 diff)

= (2C1) +                                     (2C1*11C1) +                  (2C1*7C1) +         (2C1*11C2) +           (8C2*10) +                 (8C1*11C3) + (12C5) = 2540

C = (2 sim, 2 sim, 1 diff) + (2 sim, 3 diff) + (5 diff)

=  (6C2*8C1)                + (6C1*9C3)     + (10C5) = 876

A = (5 white pawn) + (4 WP, 1 different) + (3 WP, 2 similar pieces) + (3 WP, 2 different) + (2 WP, 2 sim,1 diff) + (2 WP, 3 diff)

=  1 + (11C1) + (7C1 + 11C2) + (7C1*10C1) + (11C3) = 310

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5 replies to this topic

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Posted 15 April 2014 - 10:33 PM

Five pieces are randomly selected from a standard chess set. If it is known that at least one of these pieces is a pawn, what is the probability that at least two of these five pieces are white pawns?
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### #2 toletyrajesh

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Posted 16 April 2014 - 10:13 AM

is the answer 3874 / (32c5) ??

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### #3 Rainman

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Posted 16 April 2014 - 05:51 PM

Spoiler for

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### #4 m00li

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Posted 21 April 2014 - 03:58 AM   Best Answer

there are 8 pawns, 2 rooks, 2 knights, 2 bishops, 1 king and 1 queen in white. similarly we have 8,2,2,2,1,1 sets of unique pieces in black.

Method 1:

A = number of ways of selecting 5 out of the above

B = number of ways of selecting 5 such that no pawn is selected

C = number of ways of selecting 5 such that only 1 or 0 white pawn is selected

A = coeff of x5 in (1-x9)2(1-x3)6(1+x)4(1-x)-8 = 2540

B = coeff of x5 in (1-x3)6(1+x)4(1-x)-6 = 876

C = coeff of x5 in (1+x)(1-x9)(1-x3)6(1+x)4(1-x)-7=2230

Method 2

A = no. of ways of selecting 5 such that 2 or 3 or 4 or 5 white pawns are selected

B = no. of ways of selecting 5

C = no. of ways of selecting 5 without any pawns

B = (all 5 pieces are similar) + (4 similar and 1 different) + (3 sim, 2 sim) + (3 sim,2 different) + (2 sim, 2 sim, 1 diff) + (2 sim, 3 diff) + (5 diff)

= (2C1) +                                     (2C1*11C1) +                  (2C1*7C1) +         (2C1*11C2) +           (8C2*10) +                 (8C1*11C3) + (12C5) = 2540

C = (2 sim, 2 sim, 1 diff) + (2 sim, 3 diff) + (5 diff)

=  (6C2*8C1)                + (6C1*9C3)     + (10C5) = 876

A = (5 white pawn) + (4 WP, 1 different) + (3 WP, 2 similar pieces) + (3 WP, 2 different) + (2 WP, 2 sim,1 diff) + (2 WP, 3 diff)

=  1 + (11C1) + (7C1 + 11C2) + (7C1*10C1) + (11C3) = 310

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### #5 m00li

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Posted 21 April 2014 - 10:36 PM

there are 8 pawns, 2 rooks, 2 knights, 2 bishops, 1 king and 1 queen in white. similarly we have 8,2,2,2,1,1 sets of unique pieces in black.

Method 1:

A = number of ways of selecting 5 out of the above

B = number of ways of selecting 5 such that no pawn is selected

C = number of ways of selecting 5 such that only 1 or 0 white pawn is selected

A = coeff of x5 in (1-x9)2(1-x3)6(1+x)4(1-x)-8 = 2540

B = coeff of x5 in (1-x3)6(1+x)4(1-x)-6 = 876

C = coeff of x5 in (1+x)(1-x9)(1-x3)6(1+x)4(1-x)-7=2230

Method 2

A = no. of ways of selecting 5 such that 2 or 3 or 4 or 5 white pawns are selected

B = no. of ways of selecting 5

C = no. of ways of selecting 5 without any pawns

B = (all 5 pieces are similar) + (4 similar and 1 different) + (3 sim, 2 sim) + (3 sim,2 different) + (2 sim, 2 sim, 1 diff) + (2 sim, 3 diff) + (5 diff)

= (2C1) +                                     (2C1*11C1) +                  (2C1*7C1) +         (2C1*11C2) +           (8C2*10) +                 (8C1*11C3) + (12C5) = 2540

C = (2 sim, 2 sim, 1 diff) + (2 sim, 3 diff) + (5 diff)

=  (6C2*8C1)                + (6C1*9C3)     + (10C5) = 876

A = (5 white pawn) + (4 WP, 1 different) + (3 WP, 2 similar pieces) + (3 WP, 2 different) + (2 WP, 2 sim,1 diff) + (2 WP, 3 diff)

=  1 + (11C1) + (7C1 + 11C2) + (7C1*10C1) + (11C3) = 310

Really sorry for showing the approach here, but this was my first post immediately after joining and I didnt know I could hide spoilers or how to hide them.

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### #6 phil1882

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Posted 22 April 2014 - 03:03 PM

no biggie, in the future, just type [ spoiler ]message[ /spoiler ]but without the spaces.

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