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6.Permutation and Combination
medium
Ten persons, amongst whom are $A, B$ and $C$ to speak at a function. The number of ways in which it can be done if $A$ wants to speak before $B$ and $B$ wants to speak before $C$ is
A
$\frac{{10\;!}}{6}$
B
$3\;!\;7\;!$
C
$^{10}{P_3}\;.\;7\;!$
D
None of these
Solution
(a) For $A, B, C$ to speak in order of alphabets, $3$ places out of $10$ may be chosen first in $^{10}{C_3}$ ways.
The remaining $7$ persons can speak in $7\;!$ ways.
Hence, the number of ways in which all the $10$ person can speak is $^{10}{C_3}\;.\;7\;!\; = \frac{{10\;!}}{{3\;!}}. = \frac{{10\;!}}{6}.$
Standard 11
Mathematics
