Question

Determine the number of ways in which 6 pictures can be hung from 8 picture nails on the wall.
(A) \[{{6}^{8}}\]
(B) 20160
(C) \[{{8}^{6}}\]
(d) None of these

Answer 1

Hint: In this question, first we have to find the number of picture nails out of 8 that is needed to hang 6 picture which is 6 only then we have to determine in how many ways we can select these 6 picture nails out of 6 using combination function \[^{8}{{C}_{6}}\]. Then we have to determine how many ways 6 pictures can be hung over 6 picture nails using permutation and combination. That is, if the first picture is hanged over the first picture nail then the same picture cannot be hung over the remaining picture nails. So we have left which only 5 pictures that are to be hanged over the remaining 5 picture nail. continuing in this way we will get that the number of ways in which 6 pictures can be hung over 6 picture nails will be \[6!\]. In order to get the required number of ways, we will calculate \[^{8}{{C}_{6}}\times 6!\].

Complete step-by-step solution:
The given number if pictures are 6.
And we are given with 8 picture nails.
Since the number of picture nails required to hang 6 pictures is only 6.
Hence we will now determine in how many ways we can select 6 picture nails out of 8 using the combination function \[^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\] which is the number of ways of selecting \[r\] number items out of \[n\].
We will now calculate the number of ways in which 6 picture nails can be selected out of 8 which is given by \[^{8}{{C}_{6}}\].
\[\begin{align}
  & ^{8}{{C}_{6}}=\dfrac{8!}{6!\left( 8-6 \right)!} \\
 & =\dfrac{8\times 7\times 6!}{6!\times 2!} \\
 & =\dfrac{8\times 7}{2} \\
 & =4\times 7 \\
 & =28.......................\left( 1 \right)
\end{align}\]
Hence there are 28 number of ways of selecting 6 picture nails out of 8.
Now we will determine in how many ways we can arrange 6 pictures over these 6 picture nails.
For that, we will fir determine how many choices of pictures are these which can be hanged over the first picture nail.
There are only 6 such choices.
Now once a picture is hanged over the first picture nail, it cannot be hanged over the remaining ones. Thus the number of pictures left that can be hanged over the remaining 5 picture nails is given by
\[6-1=5\]
Again, the number of the picture left which can be hanged over the remaining 4 picture nails is given by
\[6-2=4\]
Again, the number of the picture left which can be hanged over the remaining 3 picture nails is given by
\[6-3=3\]
Again, the number of the picture left which can be hanged over the remaining 2 picture nails is given by
\[6-4=2\]
Again, the number of the picture left which can be hanged over the remaining 1 picture nails is given by
\[6-5=1\]
Hence the total number of ways in which these 6 pictures can be arranged over the selected 6 picture nails is given by
\[6\times 5\times 4\times 3\times 2\times 1=6!\]
Now we will finally determine in how many ways 6 pictures can be hung over 8 picture nails by substituting the value in equation \[\left( 1 \right)\] in \[^{8}{{C}_{6}}\times 6!\].
\[\begin{align}
  & ^{8}{{C}_{6}}\times 6!=28\times 6! \\
 & =28\times 6\times 5\times 4\times 3\times 2\times 1 \\
 & =20160
\end{align}\]
Hence there is a 20160 number of ways in which 6 pictures can be hung over 8 picture nails.
Therefore option (b) is correct.

Note: In this problem, do not determine the number ways in which 6 pictures can be hung over 8 picture nails by simply calculating \[^{8}{{C}_{6}}\] which is just 28 and the answer is wrong. Another way of solving this question is using the permutation formula which is given by \[^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}\].