Differentiation (Part-38) | S N De

In the previous article, we have discussed Short Answer Type Questions . In this article, we will discuss the solutions of Short Answer Type Questions from Ex-7A in the chapter Permutation &Combination, Class XI as given in the Chhaya Publication Book or alternatively S N Dey mathematics solution class xi book of aforementioned chapter .

 

Permutation related Long Answer Type Questions and Solutions

1. Let $~{}^nP_r~$ denote the number of permutations of n different things taken r at a time . Then, prove that

$1+1 \cdot {}^1P_1+2 \cdot {}^2P_2+3 \cdot {}^3P_3+\cdots + n \cdot {}^nP_n= {}^{n+1}P_{n+1}$

Solution.

$ {}^{i+1}P_{i+1}=(i+1)!=(i+1)i! \\ \text{or,}~~ {}^{i+1}P_{i+1} =(i+1)\times {}^iP_i=i{}^iP_i+{}^iP_i \\ \text{or,}~~{}^{i+1}P_{i+1}-{}^iP_i=i\cdot{}^iP_i \rightarrow(1)$

Clearly, $(1)~$ indicates an identity. We put $i=1,2,3, \cdots,n~$ to get,

$ {}^2P_2-{}^1P_1=1 \times {}^1P_1 \ {}^3P_3-{}^2P_2=2 \times {}^2P_2 \\~~~~ \vdots \\~~~~ {}^nP_n-{}^{n-1}P_{n-1}=(n-1) \times {}^{n-1}P_{n-1}\\~~~~ {}^{n+1}P_{n+1}-{}^nP_n=n \times {}^nP_n$

Adding we get,

$ {}^{n+1}P_{n+1}-{}^1P_1= 1 \times {}^1P_1+2 \times {}^2P_2+ \cdots +n \times {}^nP_n \\ \text{or,}~~ 1+1 \cdot {}^1P_1+2 \cdot {}^2P_2+3 \cdot {}^3P_3+\cdots + n \cdot {}^nP_n= {}^{n+1}P_{n+1}$

2. If $~ \frac{{}^nP_{r-1}}{a}=\frac{{}^nP_{r}}{b}=\frac{{}^nP_{r+1}}{c},$ prove that , $~b^2=a(b+c).$

Solution.

$ \frac1a \times {}^nP_{r-1}=\frac1b \times {}^nP_r \\ \text{or,}~~ \frac{n!}{a(n-r+1)!}=\frac{n!}{b(n-r)!} \\ \text{or,}~~ \frac{1}{a(n-r+1)}=\frac{1}{b}\rightarrow(1) $

Similarly,

$ \frac1c \times {}^nP_{r+1}=\frac1b \times {}^nP_r \\ \text{or,}~~ \frac{1}{c}=\frac{1}{b(n-r)} \\ \text{or,}~~(n-r)=\frac{c}{b}\rightarrow(2) $

Hence, from $(1)~$ and $~(2),~$ we get

$ \frac{1}{a(\frac cb +1)}=\frac 1b \\ \text{or,}~~ b=a\left(\frac cb +1\right) \\ \text{or,}~~ b^2=a(c+b) \\ \text{or,}~~ b^2=a(b+c)~~\text{(proved)} $

3.  Show that the number of permutations of  n  different things taken all at a time in which m particular things are never together is   $~n!-m!~(n-m+1)! $

Solution.

If no restriction is imposed, then $~n~$ different things can be arranged in $~n!~$ ways. Now if we consider m particular things as a single thing , then total number of things is $~(n-m+1).$

Now, these $~(n-m+1)~$ can be arranged among themselves in $~(n-m+1)! ~.$
Again, for each such arrangements, $~m~$ different things can be arranged among themselves in $~m!~$ ways.

So, the number of permutations of $~n~$ different things taken all at a time in which m particular things are together is $~m!~(n-m+1)!$

Hence, the number of permutations of $~n~$ different things taken all at a time in which m particular things are never together is $~~n!-m!~(n-m+1)!$

4. Find the number of numbers each less than $999$ and divisible by $2$ which can be formed with the digits $~2,3,4,5,6~$ and $~7~$, no digit occurring more than once in any number.

Solution.

The numbers each less than $999$ can be 1 digit numbers, 2-digit numbers and 3-digit numbers.

Now, 1-digit numbers that are divisible by $2$ which can be formed with the digits $~2,3,4,5,6~$ and $~7~$, can be $2,4,6$ and so the number of numbers in this case =3.Now, 1-digit numbers that are divisible by $2$ which can be formed with the digits $~2,3,4,5,6~$ and $~7~$, can be $2,4,6$ and so the number of numbers in this case =3.

For 2-digit numbers the unit place must be one of $~2,4,6~$ i.e. by $3$ ways , we can arrange unit place whereas the ten's place can be filled by $~{}^5P_1=5~$ ways and so in total number of arrangement is $=3 \times 5=15.$

For 3-digit numbers the unit place must be one of $~2,4,6~$ i.e. by $3$ ways we can arrange unit place ;

The tens place can be filled by $5$ ways as after selecting any one of 2,4,6 we are left with 5 choice out of 6 digits ;

The hundreds place can be filled by $4$ ways as after selecting 2 digits in two places we are left with $~6-2=4~$ choice as no digit occurring more than once in any number.

Hence, in this case, total number of arrangement is $~~3 \times 5 \times 4=60.$

So, finally total number of numbers less than $~900~$ following the given condition is $~~3+15+60=78.$

5.  Find the number of numbers less than $~1000~$ and divisible by $~5~$ which can be formed with the digits $~0,1,2,3,4,5,6,7,8,9 ~$ each digit not occurring more than once in each number.

Solution.

6.  How many numbers of four digits can be formed from the numbers $~1,2,3,4~$ ? Find the sum of all such numbers ( the digits are to be used once only ).

Solution.

Total number of numbers which can be formed by using all the four given digits only once $={}^4P_4=4!=24~$ ways.

We now need to find the sum of the digits in each place such as unit's place, tens place etc. For this let us fill up the unit's place by the digit $1$. Then the remaining 3 places can be filled up in $~3!=6~$ ways.

$\therefore~$ the sum of digits of each place is

$=1 \times 6+ 2\times 6+3 \times 6+4 \times 6=60.$

Hence, the required the sum of all such numbers ( the digits are to be used once only ) is given by

$=60(1000+100+10+1)=66660.$

7.  A library has 5 copies of one book, 4 copies each of two books, 6 copies each of three books and 1 copy each of eight books. In how many ways can all the books be arranged ?

Solution.

According to the problem, total number of books

$=5 \times 1+ 4 \times 2+ 6 \times 3+1 \times 8=39.$

Since the library has $~5~$ copies of one book, 4 copies each of two books, $~6~$ copies each of three books and $~1~$ copy each of eight books, number of ways of arranging these books is :

$=\frac{(39)!}{5! \times (4!)^2 \times (6!)^3}.$

8.  In how many ways can the letters of the word CONTACT be arranged (i) without changing the order of the vowels (ii) without changing the relative positions of vowels and consonants (iii) without changing the positions of the vowels ?

Solution.

In the given word ( CONTACT) having 7 letters, C ,T appear twice each and so the total number of arrangement is given by $~\frac{7!}{2! \times 2!}$

Now, the number of arrangement without changing the order of the vowels is given by :

$\frac 12 \times \frac{7!}{2! \times 2!}\\~~~=\frac{ 7 \times 6 \times 5!}{8}\\~~~=\frac{42 \times 120}{8}\\~~~=42 \times 15\\~~~=630 \rightarrow(i)$

Number of arrangement of consonants (5 letters) without changing the relative positions is : $~\frac{5!}{2! \times 2!}$

Number of arrangement of vowels (2 letters) without changing the relative positions is : $~2!$

So, combining both , number of arrangement of vowels and consonants without changing the relative position is :

$~\frac{5!}{2! \times 2!} \times 2! =\frac{120}{2}=60 \rightarrow(ii)$

Finally, number of arrangement of consonants (5 letters) without changing the relative positions of the vowels is :

$~\frac{5!}{2! \times 2!}=\frac{120}{4}=30 \rightarrow(iii)$

9. Four different letters are written and the corresponding addresses are correctly written on four envelopes. One letter can be replaced in one envelope. Find the number of ways so that all the letters are wrongly placed.

Solution.

Let the letters be denoted by $~A,B,C,D.$ Now the number of ways in which each of the remaining 3 letters can be placed in the envelope with the wrong address by placing the letter A in the envelope of the letter B is given below.

 

The letter A can be wrongly placed in 3 ways and for each case, each of the remaining 3 letters can be wrongly placed in different envelops in 3 ways.

So, the number of ways so that all the letters are wrongly placed is $~ 3\times 3=9.$

10. Find the total number of ways in which six '+' and four '-' signs can be arranged in a line such that no two '-' signs occur together.

Solution.

According to the problem, six '+' signs can be arranged in a row in $~\frac{6!}{6!}=1$ way.

Now, we are left with seven places (in between six "+" signs) in which four different things can be arranged in $~\frac{{}^7P_4}{4!}=35~$ ways.

So, number of ways in which six '+' and four '-' signs can be arranged in a line such that no two '-' signs occur together, is given by $~1 \times 35=35.$

11. If none of the digits $~0,1,2,3,4~$ be repeated , how many numbers of 5 significant digit can be formed with them? How many of them are divisible by $4~$?

Solution.

In order to form numbers of 5 significant digit using the given digits , the 5-digit number should not start with 0. So, For the first place, we have 4 options (out of 5 options/ digits). After placing first digit, we have 4 digits to fill up the 2nd spot. Similarly, After placing first 2 digits, for the 3rd, 4th and 5th place/spot, we have 3,2,1 options (of placing digits) respectively.

Hence, numbers of 5 significant digit can be formed with the given digits are $~ 4\times 4 \times 3 \times 2 \times 1=96.$

Second part of solution .

excerpts from the book : Full Solution of Permutation( Chhaya)

12. Find the number of positive numbers with three digits divisible by $5$ such that the digits in each number are different.

Solution.

In order to form positive numbers with three digits divisible by $5$ such that the digits in each number are different, unit place of 3-digit number must be filled in by $~0~$ and $~5.$

If we put $~0~$ in the unit place of 3-digit number, then for hundreds place and tens place we have $~9~$ and $~8~$ choices respectively. So, in total we have $~~ 9 \times 8=72~$ choices.

Now, if we put $~5~$ in the unit place of 3-digit number, then for hundreds place and tens place we have $~8~$ choices each. We have to keep in mind that after putting $~5~$ in the unit place, we have only $~8~$ choices out of $~9~$ given digits as we can not put $~0~$ in hundreds place , otherwise the 3-digit number will be converted into 2-digit number.

So, in this case, total we have $~~ 8 \times 8=64~$ choices.

Hence, combining both cases, total number of positive numbers following the required condition, is $~72+64=136.$