Questions & Answers

Question

Answers

(A) \[{{6}^{8}}\]

(B) 20160

(C) \[{{8}^{6}}\]

(d) None of these

Answer
Verified

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.