Permutations and Combinations Test 57

Given: 

The polygon given is a heptagon.

Therefore, the no. of sides of the given polygon \({n}=7\).

As we know that, no. of diagonals that can be drawn by joining the angular points of a polygon having \(n\) sides is given by: \({ }^{n} C_{2}-n=\frac{n \times(n-3)}{2}\)

No. of diagonals =\(\frac{7(7-3)}{2}\)

Therefore,

No. of diagonal of heptagon \(=14\)

The number of ways to select \(r\) things out of \(n\) things is given by \({ }^{n} C_{r}\)

\({ }^{{n}} {C}_{{r}}=\frac{{n} !}{({n}-{r}) ! \times({r}) !}=\frac{{n} \times({n}-1) \times \ldots({n}-{r}+1)}{{r} !}\)

\({ }^{{n}} {C}_{{r}}={ }^{{n}} {C}_{{n}-{r}}\)

Here, the number of points \(=12\)

Octagon is formed by 8 points.

\(\therefore\) Number of octagon \(=\) Number of ways of selecting 8 points out of 12

\(={ }^{12} {C}_{8}\)

\(={ }^{12} {C}_{4} \quad\left(\because{ }^{{n}} {C}_{{r}}={ }^{{n}} {C}_{{n}-{r}}\right)\)

\(=\frac{12 !}{4 ! 8 !}\) \(=495\)

We know that:

The number of ways to select \(r\) things out of \(n\) things is given by \({ }^{{n}} C_{{r}}\).

\({ }^{{n}} {C}_{{r}}=\frac{{n} !}{({n}-{r}) ! \times({r}) !}=\frac{{n} \times({n}-1) \times \ldots({n}-{r}+1)}{{r} !}\)

Given that:

There are 13 points in a plane of which 5 are collinear.

We know that:

To form a line we have to select two points out of 13 points.

\(\therefore\) Number of lines\(={ }^{13} {C}_{2}\)

\(=\frac{13 \times 12}{2 \times 1}\)

\(=78\)

Also number of lines out of 5 points\(={ }^{5} {C}_{2}\)

\(=\frac{5 \times 4}{2 \times 1}\)

\(=10\)

But, these 5 points are collinear, and only one line can be formed out of these points.

∴ The total number of straight lines obtained by joining these points in pairs.

\(=78-10+1\)\(\quad\)(We add 1, as one line can be obtained out of 5 collinear points).

\(=69\)

We know that:

To arrange \(n\) things in an order of a number of objects taken \(r\) things is:

\( { }^{n} P_{r}=\frac{n!}{(n-r)!}\)

or, \( { }^{n} P_{r}=\frac{n \times (n-1)\times(n-2)\times \cdots \times (n-r)\times \cdots \times3\times2\times1}{(n-r)!}\)

Given:

Number of signals that can be sent using 1 flag is:

\({ }^{4} \mathrm{P}_{1}=\frac{4!}{(4-1)!}=4\)

Number of signals that can be sent using 2 flags is:

\({ }^{4} \mathrm{P}_{2}=\frac{4!}{(4-2)!}=12\)

Number of signals that can be sent using 3 flags is:

\({ }^{4} P_{3}=\frac{4!}{(4-3)!}=24\)

Number of signals that can be sent using 4 flags is:

\({ }^{4} P_{4}=\frac{4!}{(4-4)!}=24\)

\(\therefore\) Total number of signals that can be sent at a time \(=4+12+24+24=64\)

The given problem can be written as,

\({ }^{n} C_{r}+{ }^{n} C_{r+1}+{ }^{n} C_{r+1}+{ }^{n} C_{r+2}\)

We know that:

\({ }^{n} C_{r}=\frac{n !}{r !(n-r) !}\)

Therefore,

\({ }^{n} C_{r}+{ }^{n} C_{r+1}+{ }^{n} C_{r+1}+{ }^{n} C_{r+2}\)

\(=\frac{{n} !}{{r} !({n}-{r})!}+\frac{{n} !}{({r}+1) !({n}-{r}-1) ! }+\frac{{n} !}{({r}+1) !({n}-{r}-1) ! }+\frac{{n} !}{({r}+2) !({n}-{r}-2) ! }\)

\(=\frac{{n} !}{{r} !({n}-{r}-1) ! }\left[\frac{1}{({n}-{r})}+\frac{1}{({r}+1)}\right]+\frac{{n} !}{({r}+1) !({n}-{r}-2) ! }\left[\frac{1}{({n}-{r}-1)}+\frac{1}{({r}+2)}\right]\)

\(=\frac{{n} !}{{r} !({n}-{r}-1) ! }\left[\frac{{r}+1+{n}-{r}}{({n}-{r})({r}+1)}\right]+\frac{{n} !}{({r}+1) !({n}-{r}-2) ! }\left[\frac{{r}+2+{n}-{r}-1}{({n}-{r}-1)({r}+2)}\right]\)

\(=\frac{{n} ! \times({n}+1)}{{r} ! \times({r}+1) \times({n}-{r}-1) \times({n}-{r})}+\frac{{n} !}{({r}+1) !({n}-{r}-2) !} \times \frac{({n}+1)}{({n}-{r}-1)({r}+2)}\)

\(=\frac{({n}+1) !}{({r}+1) !({n}-{r}) !}+\frac{{n} ! \times(\mathbf{n}+1)}{({r}+2)({r}+1) \times({n}-{r}-1)({n}-{r}-2)}\)

\(=\frac{({n}+1) !}{({r}+1) !({n}-{r}) !}+\frac{({n}+1) !}{({r}+2) !({n}-{r}-1) !}\)

\(=\frac{({n}+1) !}{({r}+1) !({n}-{r}-1) !}\left[\frac{1}{{n}-{r}}+\frac{1}{{r}+2}\right]\)

\(=\frac{({n}+1) !}{({r}+1) !({n}-{r}-1) !}\left[\frac{{n}-{r}+{r}+2}{({r}+2)({n}-{r})}\right]\)

\(=\frac{({n}+2)({n}+1) !}{({r}+2)({r}+1) !({n}-{r})({n}-{r}-1) !}\)

\(=\frac{({n}+2) !}{({r}+2) !({n}-{r}) !}\)

\(=^{{n}+2} {C}_{{r}+2}\)

\(=C(n+2, r+2)\)

\({ }^{2 {n}} {C}_{3}:^{{n}} {C}_{3}=17: 2\)

\(\Rightarrow \frac{2 {n}(2 {n}-1)(2 {n}-2)}{3 !}: \frac{{n}({n}-1)({n}-2)}{3 !}=17: 2\)

\(\Rightarrow \frac{2 {n}(2 {n}-1)(2 {n}-2)}{{n}({n}-1)({n}-2)}=\frac{17}{2}\)

\(\Rightarrow \frac{4({n}-1)(2 {n}-1)}{({n}-1)({n}-2)}=\frac{17}{2}\)

\(\Rightarrow \frac{4(2 {n}-1)}{({n}-2)}=\frac{17}{2}\)

\(\Rightarrow 16 {n}-8=17 {n}-34\) \(\Rightarrow {n}=26\)

Given that:

4 boys and 4 girls can be arranged in a row so that no two girls and no two boys are together.

It means they can sit alternately.

Case-I: 1st person in the row is a boy.

\({B}_{1} {G}_{1} {~B}_{2} {G}_{2} {~B}_{3} {G}_{3} {~B}_{4} {G}_{4}\)

The no. of ways in which the boys can be rearranged among themselves is \(4 !\).

Similarly, the no. of ways in which the girls can be rearranged among themselves is \(4 !\)

So, the total number of ways in this case \(=4 ! \times 4 !=(4 !)^{2}\)

Case-II: 1st person in the row is a girl.

\({G}_{1} {~B}_{1} {G}_{2} {~B}_{2} {G}_{3} {~B}_{3} {G}_{4} {~B}_{4}\)

The no. of ways in which the girls can be rearranged among themselves is \(4 ! .\)

Similarly, the no. of ways in which the boys can be rearranged among themselves is \(4 !\)

So, the total number of ways in this case \(=4 ! \times 4 !=(4 !)^{2}\)

Therefore, Total Number of ways \(=(4 !)^{2}+(4 !)^{2}=2(4 !)^{2}\)

Given:

\(C(30, n-3)=C(30, n+3)\)

We know that:

If \({ }^{n} C_{x}={ }^{n} C_{y}\), then

\(x=y\)

or, \(x+y=n\)

Therefore,

\({ }^{30} {C}_{n-3}={ }^{30} {C}_{n}+3\)

\({n}-3+{n}+3=30\)

\(2 {n}=30\)

\({n}=15\)

Given:

\({ }^{2 n} P_{3}=100^{n} P_{2}\)

We know that:

\(P_{r}=\frac{n !}{(n-r) !}\)

\(n !=1 \times 2 \times 3 \times \ldots \times n\)

\(0 !=1\)

Therefore,

\({ }^{2 n} P_{3}=100^{n} P_{2}\)

\(\Rightarrow \frac{2 n !}{(2 n-3) !}=100 \frac{n !}{(n-2) !}\)

Expanding the expression on both sides of the equation, we get:

\(\Rightarrow(2 n)(2 n-1)(2 n-2)=100(n)(n-1)\)

\(\Rightarrow 2 n-1=25, n \neq 1\)

\(\Rightarrow n=13\)

We know that:

\({ }^{n} C_{r}=\frac{n !}{r !(n-r) !}\)

A particular player should not be included,

We have to select 5 players from (8 - 1) = 7 players.

Therefore, required number of ways,

\(={ }^{7} {C}_{5}\)

\(=\frac{7 !}{5 !(7-5) !}=21\)