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Permutations and Combinations
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Ten persons are to speak in a meeting. The no. of ways in which this can be arranged if A wants to speak before B and B wants to speak before C, is ?
jeemain
math
class11
ch7
permutations-and-combinations
permutations
medium
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asked
Aug 27, 2013
by
rvidyagovindarajan_1
edited
Aug 6, 2014
by
sharmaaparna1
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No. of arrangement of n different things out of which 'm' are of one category = $\large\frac{n!}{m!}$
The no. of arrangement of 10 speakers without any condition = $(10)!$
A,B, C can speak any order = $3!$ ways.
But A,B,C has a fixed arrangement.
$\therefore$ The required arrangement = $\large\frac{(10)!}{6}$
answered
Aug 27, 2013
by
rvidyagovindarajan_1
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