Permutations and Combinations Test 59

Given that,

\({ }^{22} {C}_{{m}}={ }^{22} {C}_{{n}}\)

Here,\(n=22\)

We know that,

If \({ }^{{n}} {C}_{{x}}={ }^{{n}} {C}_{{y}}\) and \({x} \neq {y}\), then

\({n}={x}+{y}\)

\(\therefore {m}+{n}=22\)

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:

We have to select 2 persons out of 76 for handshakes.

\(\therefore\) Number of handshakes\(={ }^{76} {C}_{2}\)

\(=\frac{76 \times 75}{2 \times 1}\)

\(=38 \times 75\)

\(=2850\)

Therefore,\(2850\) handshakes are possible.

Given word is : INDIANARMY

Total 10 letters are there, in which \({A}, {I}\) and \({N}\) are repeating twice.

We know that:

Number of Permutations of '\({n}\)' things taken '\({r}\)' at a time:

\(p(n, r)=\frac{n !}{(n-r) !}\)

Number of Permutations of ‘\(n\)’ objects where there are \(n_1\) repeated items, \(n_2\) repeated items, \(n_k\) repeated items taken ‘\(r\)’ at a time:

\(p(n, r)=\frac{n !}{n_{1} ! n_{2} ! n_{3} ! \ldots n_{k} !}\)

Therefore,

Number of different permutations \(=\frac{10 !}{2 ! \times 2 ! \times 2 !}\)

\(=\frac{10 !}{(2 !)^{3}}\)

Therefore, the permutations can be made out of the letters of the word 'INDIANARMY' will be \(\frac{10 !}{(2 !)^{3}}\).

The number of ways to select \({r}\) things out of \({n}\) given things wherein \({r} \leq {n}\) is given by:

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

From 2 wicket-keepers we have to select 1, so, number of ways of selecting \(={ }^{2} {C}_{1}\)

From 5 bowlers we have to select 3 bowlers, so, the number of ways of selecting \(={ }^{5} {C}_{3}\)

From 8 batsmen we have to select 7 batsmen, so, the number of ways of selecting \(={ }^{8} {C}_{7}\)

Therefore,

Total required number of ways

\(={ }^{2} C_{1} \times{ }^{5} {C}_{3} \times{ }^{8} {C}_{7}\)

\(=2 \times 10 \times 8\)

\(=160\) Ways

We know that:

Number of arrangement of \(n\) different objects is given by n!

\({n} !={n} \times({n}-1) \times \ldots \times 1\)

The word VOWELS contains 6 letters, in which all the different letters occur only once.

Now, after fixing \({O}\) in first place and \({E}\) in the last place, we have only 4 letters.

\(\therefore\) Number of words \(=4 !=24\).

We know that:

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

Given that:

\({ }^{10} {P}_{{r}}=5040\)

\(\Rightarrow{ }^{10} {P}_{{r}}=10 \times 504\)

\(\Rightarrow{ }^{10} {P}_{{r}}=10 \times 9 \times 56\)

\(\Rightarrow{ }^{10} {P}_{{r}}=10 \times 9 \times 8 \times 7\)

\(\Rightarrow{ }^{10} {P}_{{r}}=10 \times(10-1) \times(10-2) \times(10-4+1)\)

On comparing the above expression with \({ }^{{n}} {P}_{{r}}={n} \times({n}-1) \times \ldots \ldots \times({n}-{r}+1)\), we get:

\(r=4\)

We know that:

The number of ways to select \({r}\) things out of \({n}\) given things wherein \({r} \leq {n}\) is given by:

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

We have to select 3 men out of 7 men, so, the number of ways of selection \(={ }^{7} C_{3}\)

Now, we have to select 3 women out of 5 women, so, the number of ways of selection \(={ }^{5} {C}_{2}\)

Therefore,

Total required no. of ways

\(={ }^{7} {C}_{3} \times{ }^{5} {C}_{2}\)

\( =\frac{7 !}{3 ! \times(7-3) !} \times \frac{5 !}{2 ! \times(5-2) !}\)

\(=\frac{7 \times 6\times 5}{3 \times 2 \times 1} \times \frac{5 \times 4}{2 \times 1}\)

\(=\frac{7 \times 6\times 5}{3 \times 2 \times 1} \times \frac{5 \times 4}{2 \times 1}\)

\(=350\)

We know that:

The number of permutations of n objects taken r at a time is given by:

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

where,

n = number of objects

r = number of positions

There can be only one arrangement in which boys and girls can sit.

B - G - B - G - B - G - B - G - B

But the position of individual boys and girls can be changed.

So, there are 5 boys and 5 slots for them that means they can be arranged in \({ }^{5} {P}_{5}\) or 5! ways,

Similarly girls have 4 slots, so they can be arranged in \({ }^{4} {P}_{4}\) or \(4 !\) ways.

Now, one more thing to consider is that out of \(5 !\) arrangements of boys, with each arrangement girls can be arranged in 4! different positions.

So, to get the total number of boys and girls arrangements together we would need to multiply the arrangements of both boys and girls.

So, the desired answer would be \(={ }^{5} {P}_{5} \times{ }^{4} {P}_{4}=5 ! \times 4 !=120 \times 24=2880\) ways.

Given that:

\({ }^{n} C_{15}={ }^{n} C_{8}\)

As we know that,

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

\(x+y=n\)

Therefore,

\(n=15+8=23\)

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:

Here, we have 7 boys and 4 girls out of which we have to select an equal number of boys and girls to form a team of 6 (i.e., 3 boys and 3 girls), that is,

\(\therefore\) Required number of ways \(={ }^{7} {C}_{3} \times{ }^{4} {C}_{3}\)

\(=\frac{7 \times 6 \times 5}{3 !} \times \frac{4 \times 3 \times 2}{3 !}\)

\(=\frac{7 \times 6 \times 5}{3 \times 2} \times \frac{4 \times 3 \times 2}{3 \times 2}\)

\(=35 \times 4\)

\(=140\)