Permutations and Combinations Test 58

If we fix a girl, then the rest 4 girls can be arranged in the ways \(=4 !\)

There is 1 position between each pair of girl, which means only 5 positions are there.

Ways to arrange 5 boys in those 5 position will be \(=5!\)

\(\therefore\) The total number of ways, \(N=5 ! \times 4!\)

\(N=120 \times 24\)

\(N=2880\) ways

We know that:

Suppose a set of \(n\) objects has \(n_{1}\) of one kind of object, \(n_{2}\) of a second kind, \(n_{3}\) of a third kind, and,

So, on with \(n=n_{1}+n_{2}+n_{3}+\ldots+n_{k}\) then the number of distinguishable permutations of the \(n\) objects is:

\(=\frac{n !}{n_{1} ! \times n_{2} ! \times n_{3} ! \ldots \ldots . n_{k} !}\)

Given: Three a’s and two b’s

Given that:

\({ }^{{n}} {P}_{{r}}=2760,{ }^{{n}} {C}_{{r}}=23\)

We know that,

\({ }^{{n}} {C}_{{r}}=\frac{{ }^{{n}} {P}_{{r}}}{{r} !}\)

\(\Rightarrow 23=\frac{2760}{r !}\)

\(\Rightarrow {r} !=\frac{2760}{23}=120\)

\(\Rightarrow {r} !=5 \times 24\)

\(\Rightarrow {r} !=5 \times 4 \times 6\)

\(\Rightarrow {r} !=5 \times 4 \times 3 \times 2 \times 1\)

\(\Rightarrow {r!}=5!\)

\(\therefore r=5\)

We know that:

Suppose a set of \(n\) objects has \(n_{1}\) of one kind of object, \(n_{2}\) of a second kind, \(n_{3}\) of a third kind, and so, on with \(n=n_{1}+n_{2}+n_{3}+\ldots+n_{k}\), then the number of distinguishable permutations of the \(n\) objects is:

\(=\frac{n !}{n_{1} ! \times n_{2} ! \times n_{3} ! \ldots \ldots . . n_{k} !}\)

If there are m ways to choose an object one and n ways to choose an object two then the number of ways of selecting objects one and two are given by m × n.

If there are m ways to choose an object one and n ways to choose an object two then the number of ways of selecting objects one or two is given by m + n.

Given:

Total no. of toys = 5

Total no. of children = 3

Each should get one toy.

Selection can be done as follows: (2, 2, 1) or (1, 1, 3)

\(={ }^{5} {C}_{2} \times{ }^{3} {C}_{2} \times{ }^{1} {C}_{1} \times \frac{3 !}{2 !}+{ }^{5} {C}_{1} \times{ }^{4} {C}_{1} \times{ }^{3} {C}_{3} \times \frac{3 !}{2 !}\)

\(=(10 \times 3 \times 1 \times 3)+(5 \times 4 \times 1 \times 3)\)

\(=90+60\)

\(=150\)

Given word is : ASHUTOSH

Total 8 letters are there, in which letter \({S}\) and \({H}\) are repeated 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,The number of arrangements will be:

\(p(8,2)=\frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 !}{(2 \times 1) 2 !}\)\(=10080\)

Therefore, the number of arrangements of letters in the word ASHUTOSH will be \(10080\)

We know that:

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

Given that:

\(5 \times{ }^{{n}} {P}_{4}=7 \times{ }^{({n}-1)} {P}_{4}\)

\(\Rightarrow 5 \times n \times(n-1) \times(n-2) \times(n-3)=7 \times(n-1) \times(n-2) \times(n-3) \times(n-4)\)

\(\Rightarrow 5 \times n=7 \times(n-4)\)

\(\Rightarrow 5 n=7 n-28\)

\(\Rightarrow 2 n=28\)

\(\Rightarrow {n}=14\)

The word ALLAHABAD contains 9 letters, in which A occur 4 times, L occurs twice and the rest of the letters occur only once.

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 words formed by the word ALLAHABAD using all the letters \(=\frac{9 !}{4 ! \times 2 !}\)

\(=\frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 !}{4 ! \times 2}\) \(=7560\)

Now, let us take both L together and consider (LL) as 1 letter.

Then, we will have to arrange 8 letters, in which A occurs 4 times and the rest of the letters occur only once.

So, the number of words having both \(L\) together will be \(=\frac{8 !}{4 !}\)

\(=\frac{8 \times 7 \times 6 \times 5 \times 4 !}{4 !}\) \(=1680\)

Therefore, the number of words with both L not occurring together will be \(= 7560 - 1680\) \(= 5880\)

Given that:

All boys are to sit together.

So, all boys can be considered as a single group.

\(\therefore\) Total no. of students \(=4\) Girls \(+1\) Group \(=5\)

Since, this is a cyclic permutation.

We know that, for cyclic permutation:

\({}^nP_r=(n-1)!\)

Therefore,

The number of ways of arranging 5 students in a round table is \((5-1) !=4 !\)

Now, no. of the ways of arranging 5 boys is \(5 !\).

Therefore, the total number of ways \(=4 ! 5 !\)

Given that:

\({ }^{9} P_{5}+5 .{ }^{9} P_{4}={ }^{10} P_{r}\)

We know that:

Number of Permutations of ‘n’ things taken ‘r’ at a time:

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

\(\frac{9 !}{4 !}+5 \cdot \frac{9 !}{5 !}=\frac{10 \times 9 !}{(10-r) !}\)

\(9 !\left[\frac{1}{4 !}+\frac{5}{5 !}\right]=\frac{10 \times 9 !}{(10-r) !}\)

\(\left[\frac{1}{4 !}+\frac{5}{5 \times 4 !}\right]=\frac{10}{(10-r) !}\)

\(\Rightarrow \frac{2}{4 !}=\frac{10}{(10-r) !}\)

\(\Rightarrow \frac{1}{4 !}=\frac{5}{(10-r) !}\)

\(\Rightarrow(10-r) !=5 \times 4 !\)

\(\Rightarrow(10-r) !=5 !\)

= 5

Given that:

There are 20 cricket players, out of which 5 players can bowl.

We have to make a team of 11 players so to include 4 bowlers.

So, we select 4 bowlers out of 5 players and the remaining 7 players must be selected from 15 players, i.e.,

Total ways \(={ }^{5} {C}_{4} \times{ }^{15} {C}_{7}\)