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Permutations and Combinations
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There are $10$ white and $10$ black balls each set numbered 1,2,3......10. The no. of ways in which they can be arranged so that no two adjacent balls are of same colour is ?
jeemain
math
class11
ch7
permutations-and-combinations
fundamental-principle-of-counting
medium
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asked
Sep 5, 2013
by
rvidyagovindarajan_1
edited
Aug 6, 2014
by
sharmaaparna1
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1 Answer
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Case (i)
First let us place 10 white balls in (10!) ways.
Then 10 black balls are to be placed in between the white balls in (10!) ways.
Case (ii)
First place 10 black balls in 10! ways and
then place the white balls in between the black balls in 10! ways.
$\therefore $ The required no. of arrangements $= 10!.10!+10!.10!=2\times (10!)^2$
answered
Sep 5, 2013
by
rvidyagovindarajan_1
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