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A local post office is to send $M$ telegrams which are distributed at random over $N$ communication channels $(N > M)$ . Each telegram is sent over any channel with equal probability. The chance that not more than one telegram will be sent over each channel is –
$
  A)\,\dfrac{{{}^N{c_M}\,M!}}{{{N^M}}} \\
  B)\,\dfrac{{{}^N{c_M}\,N!}}{{{M^N}}} \\
  C)\,1 - \dfrac{{{}^N{c_M}\,M!}}{{{M^N}}} \\
  D)\,1 - \dfrac{{{}^N{c_M}\,N!}}{{{N^M}}} \\
 $

Answer
VerifiedVerified
497.1k+ views
Hint: There total no. of telegram is $M$ and number of channels is $N{\text{ and (}}N > M)$
So total no of ways $ = {N^M}$
No. of ways of choosing $M$ telegrams to $N$ channels is equal to ${}^N{C_M}$
Now you need to find the chance that not more than one telegram will be sent over each channel.

Complete step-by-step solution:
Here we are given that the post office is to send $M$ telegrams which are further distributed over $N$ no. of communication channels. And given that $N > M$
So it is given –
No of telegrams to be distributed $ = M$
No of communication channel $ = N$
So total no of ways of distributing $M$ telegrams to $N$ channels is $ = {N^M}$
So total no of ways ways $ = {N^M}$
Now first of all $M$ telegram chooses from the $N$ channel in ${}^N{c_M}$ no of ways and number of ways of sending $M$ telegrams through $N$ channels where $N > M$$ = M!$
So we got that the favorable ways $ = {}^N{c_M}M!$
Hence we also know the total no of ways is $ = {N^M}$
So the probability that not more than one telegram will be sent over each channel is given by $\dfrac{{{}^N{c_M}\,M!}}{{{N^M}}}$

So option A is correct.

Note: There total no. of telegrams is $M$ and number of channels is $N\,\,\& \,\,(N > M)$. So total no of ways of distributing $M$ telegrams to $N$ channels is $ = {N^M}$ And according to the question favorable case is equal to $ = {}^N{c_M}M!$ as $M$ telegrams first choose $N$ channels then $M$ telegrams pass through $N$ channels in ${}^N{c_M}M!$ no. of ways.