Chapter 6: Permutations and Combinations

Ans:

We have to form a $3$-digit numbers using digits $1-9$. That means, the order of the digits matter.

Hence, there will be as many $3$-digit numbers as there are permutations of $9$ digits taken $3$ at a time.

So, required number of $3$-digit numbers 

$={{\ }^{9}}{{P}_{3}}$

 $\Rightarrow \frac{9!}{(9-3)!}=\frac{9!}{6!}$

$\Rightarrow \frac{9\times 8\times 7\times 6!}{6!}$

$\Rightarrow 9\times 8\times 7=504$

Hence, there are $504$  $3$-digit numbers formed.

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The thousands place of the $4$-digit number is to be filled with any of the digits from $1$ to $9$ as the digit $0$ cannot be included. Therefore, the number of ways in which thousands place can be filled is $9$. The hundreds, tens, and units place can be filled by any of the digits from $0$ to $9$. However, the digits cannot be repeated in the $4$-digit numbers and thousands place is already occupied with a digit. The hundreds, tens, and units place is to be filled by the remaining $9$-digits. Therefore, there will be as many such $3$-digit numbers as there are permutations of $9$ different digits taken $3$ at a time.

Number of such $3$-digit numbers $={{\ }^{9}}{{P}_{3}}$

$\Rightarrow \frac{9!}{(9-3)!}=\frac{9!}{6!}$

$\Rightarrow \frac{9\times 8\times 7\times 6!}{6!}$

$\Rightarrow 9\times 8\times 7=504$

Thus, by multiplication principle, the required number of $4$-digit numbers are $9\times 504=4536$.

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We have to form $3$-digit even numbers using the given six digits, \[1,\text{ }2,\text{ }3,\text{ }4,\text{ }6,\] ,and $7$, without repeating the digits. Then, units can be filled in $3$ ways by any of the digits, $2$, $4$ or $6$. Since the digits cannot be repeated in the $3$-digit numbers and units place is already occupied with a digit (which is even), the hundreds and tens place is to be filled by the remaining $5$-digits. Therefore, the number of ways in which hundreds and tens place can be filled with the remaining $5$-digits is the permutation of $5$ different digits taken $2$ at a time.

So, the number of ways of filling hundreds and tens place

${{\Rightarrow }^{5}}{{P}_{2}}$

$\Rightarrow \frac{5!}{(5-2)!}=\frac{5!}{3!}$

$\Rightarrow \frac{5\times 4\times 3!}{3!}$

$\Rightarrow 5\times 4=20$

Thus, by multiplication principle, the required number of $3$-digit numbers are $3\times 20=60$.

Ans:

We have to form $4$-digit numbers using the digits, $1,\ 2,\ 3,\ 4,$ and $5$. 

So, there will be as many $4$-digit numbers as there are permutations of $5$ different digits taken $4$ at a time.

That is, required number of $4$-digit numbers

${{\Rightarrow }^{5}}{{P}_{4}}$

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

$\Rightarrow 1\times 2\times 3\times 4\times 5$

$\Rightarrow 120$

Among the $4$-digit numbers formed by using the digits, $1,\ 2,\ 3,\ 4,\ 5$ even numbers end with either $2$ or $4$. The number of ways in which units place is filled with digits is $2$. Since the digits are not repeated and the units place is already occupied with a digit (which is even), the remaining places are to be filled by the remaining $4$-digits. Therefore, the number of ways in which the remaining places can be filled is the permutation of $4$ different digits taken $3$ at time. 

Number of ways of filling the remaining places

${{\Rightarrow }^{4}}{{P}_{3}}=\frac{4!}{(4-3)!}$

$\Rightarrow 4!$

$\Rightarrow 4\times 3\times 2\times 1=24$

Thus, by multiplication principle, the required number of even numbers is $24\times 2=48$.

Ans:

We have to choose a chairman and a vice chairman from a committee of $8$ persons in such a way that one person cannot hold more than one position. Here, the number of ways of choosing a chairman and a vice chairman is the permutation of $8$ different objects taken $2$ at a time. 

Thus, required number of ways

${{\Rightarrow }^{8}}{{P}_{2}}=\frac{8!}{(8-2)!}$

$\Rightarrow \frac{8\times 7\times 6!}{6!}=8\times 7$

$\Rightarrow 56$

Ans:

Here,

$^{n-1}{{P}_{3}}{{:}^{n}}{{P}_{4}}=1:9$

$\Rightarrow \frac{^{n-1}{{P}_{3}}}{^{n}{{P}_{4}}}=\frac{1}{9}$

$\Rightarrow \frac{\left[ \frac{(n-1)!}{(n-1-3)!} \right]}{\left[ \frac{n!}{(n-4)!} \right]}=\frac{1}{9}$

$\Rightarrow \frac{(n-1)!}{(n-4)!}\times \frac{(n-4)!}{n!}=\frac{1}{9}$

$\Rightarrow \frac{(n-1)!}{n\times (n-1)!}=\frac{1}{9}$

\[\Rightarrow \frac{1}{n}=\frac{1}{9}\]

Hence, $n=9$.

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$\Rightarrow \frac{5!}{(5-r)!}=2\times \frac{6!}{(6-r+1)!}$

$\Rightarrow \frac{5!}{(5-r)!}=\frac{2\times 6\times 5!}{(7-r)(6-r)(5-r)!}$

$\Rightarrow (7-r)(6-r)=12$

$\Rightarrow 42-6r-7r+{{r}^{2}}=12$

$\Rightarrow r(r-3)-10(r-3)=0$

$\Rightarrow r=3\text{ or }r=10$

It is known that $^{n}{{P}_{r}}=\frac{n!}{(n-r)!}$, where $0\le r\le n$.

$\Rightarrow 0\le r\le 5$

Hence, $r=3$.

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$\Rightarrow \frac{5!}{(5-r)!}=\frac{6!}{(6-r+1)!}$

$\Rightarrow \frac{5!}{(5-r)!}=\frac{6\times 5!}{(7-r)(6-r)(5-r)!}$

$\Rightarrow (7-r)(6-r)=6$

$\Rightarrow 42-6r-7r+{{r}^{2}}=6$

$\Rightarrow r(r-4)-9(r-4)=0$

It is known that $^{n}{{P}_{r}}=\frac{n!}{(n-r)!}$, where $0\le r\le n$.

$\Rightarrow 0\le r\le 5$

Hence, $r=4$.

Ans:

There are $8$ different letters in the word EQUATION. Therefore, the number of words that can be formed using all the letters of the word EQUATION, using each letter exactly once, is the number of permutations of $8$ different objects taken $8$ at a time, which is $^{8}{{P}_{8}}=8!$.

Thus, required number of words that can be formed is

$8!=40320$

Ans:

There are 6 different letters in the word MONDAY.

Number of $4$-letter words that can be formed from the letters of the word MONDAY, without repetition of letters, is the number of permutations of $6$ different objects taken $4$ at a time, which is $^{6}{{P}_{4}}$.

Thus, required number of words that can be formed using $4$ letters at a time are $^{6}{{P}_{4}}=\frac{6!}{(6-4)!}$

\[\Rightarrow \frac{6\times 5\times 4\times 3\times 2!}{2!}=6\times 5\times 4\times 3\]

$\Rightarrow 360$

Ans:

Number of words that can be formed by using all the letters of the word MONDAY at a time is the number of permutations of $6$ different objects taken $6$ at a time, which is $^{6}{{P}_{6}}=6!$.

Thus, required number of words that can be formed when all letters are used at a time

$6!=1\times 2\times 3\times 4\times 5\times 6$.

$\Rightarrow 720$

Ans:

In the given word, there are $2$ different vowels, which have to occupy the rightmost place of the words formed. This can be done only in $2$ ways. Since the letters cannot be repeated and the rightmost place is already occupied with a letter (which is a vowel), the remaining five places are to be filled by the remaining $5$ letters. This can be done in \[5!\] ways.

Thus, in this case, required number of words that can be formed is \[5!\times 2=120\times 2\]

$\Rightarrow 240$

Ans: 

In the given word MISSISSIPPI, I appear $4$ times, S appears $4$ times, P appears $2$ times, and M appears just once. Therefore, number of distinct permutations of the letters in the given word

$\Rightarrow \frac{11!}{4!4!2!}$

$\Rightarrow \frac{11\times 10\times 9\times 8\times 7\times 6\times 5\times 4!}{4!\times 4\times 3\times 2\times 1\times 2\times 1}$

$\Rightarrow \frac{11\times 10\times 9\times 8\times 7\times 6\times 5}{4\times 3\times 2\times 1\times 2\times 1}$

$\Rightarrow 34650$

There are $4$ I’s in the given word. When they occur together, they are treated as single object [IIII] for the time being. This single object together with the remaining $7$ objects will account for $8$ objects.

These $8$ objects in which there are $4$ S’s, and $2$ P’s can be arranged in $\frac{8!}{4!2!}$ ways that means, $840$ ways.

Number of arrangements where all I’s occurred together is in $840$ ways.

Thus, number of distinct permutations of the letters in MISSISSIPPI in which four IS do not come together \[34650-840=33810\].