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Question: There is n number of seats and m number of people having to be seated in those n seats. How many pos...

There is n number of seats and m number of people having to be seated in those n seats. How many possible arrangements are possible, if we are given m is less than n.

Explanation

Solution

Now we know that the number of ways in which r objects can be placed in n places is given by nCr^{n}{{C}_{r}} . Now after placing m people we can arrange them in m! ways. Hence the total number of arrangement is nCrm!^{n}{{C}_{r}}m! .

Complete step-by-step solution:
Now in a given room, we gave n number of seats.
Now we want to place n people in those n seats. Hence to do so we will first select m seats from n.
Now we know that the number of ways of selecting r objects from n objects is nCr^{n}{{C}_{r}} .
Where nCr=n!(n−r)!r!^{n}{{C}_{r}}=\dfrac{n!}{(n-r)!r!} and a!=a×(a−1)×(a−2)×....×(2)×1a!=a\times (a-1)\times (a-2)\times ....\times (2)\times 1
Hence we know that the number of ways of selecting m seats from n seats is given by.
nCm^{n}{{C}_{m}} . …………………… (1)
To understand this with an example
Consider 3 chairs numbered 1, 2, 3.
Now we want to select 2 chairs. Hence we can choose (1, 2), (2, 3), (1, 3).
Hence we have 3 selections
Similarly if we choose it by formula we have 3C2=3!2!(3−2)!=3×22×1(1)!=3^{3}{{C}_{2}}=\dfrac{3!}{2!(3-2)!}=\dfrac{3\times 2}{2\times 1(1)!}=3
Now once we have selected m seats we will place m people on these seats.
This is the same as arranging m objects. Now we know that the total arrangements of m objects is given by m!
Hence the number of ways in which we can place m people on m seats is m! ………………………(2)
Hence from equation (1) and equation (2) we get the number of total arrangement is
nCmm!^{n}{{C}_{m}}m!
Now we know that nCr=n!(n−r)!r!^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}
Hence we have
nCmm!=n!(n−m)!m!×m! nCmm!=n!(n−m)! \begin{aligned} & ^{n}{{C}_{m}}m!=\dfrac{n!}{\left( n-m \right)!m!}\times m! \\\ & ^{n}{{C}_{m}}m!=\dfrac{n!}{\left( n-m \right)!} \\\ \end{aligned}
Hence we have the total number of possible arrangements is n!(n−m)!\dfrac{n!}{\left( n-m \right)!}

Note: Now we have that number of arrangements of r objects in n places is given by nPr^{n}{{P}_{r}} where nPr=n!(n−r)!^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!} . hence using this we can directly find the number of ways to place m people in n chairs.