Exercise 4.2 - Chapter 4 Combinatorics and Mathematical Induction 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Ex 4.2
Question 1.
If $(\mathrm{n}-1) \mathrm{P}_3:{ }^n \mathrm{P}_4$, find $\mathrm{n}$ :
Solution:
$
\begin{aligned}
& \text { Given } \frac{(n-1) \mathrm{P}_3}{{ }^n P_4}=\frac{1}{10} \\
& \Rightarrow \frac{(n-1)(n-2)(n-3)}{n(n-1)(n-2)(n-3)}=\frac{1}{10} \text {; (i.e) } \frac{1}{n}=\frac{1}{10} \Rightarrow n=10
\end{aligned}
$
Question 2.
If ${ }^{10} \mathrm{P}_{\mathrm{r}-1}=2 \times{ }^6 \mathrm{P}_{\mathrm{r}}$, find $\mathrm{r}$.
Solution:
$
\begin{aligned}
{ }^{10} \mathrm{P}_{r-1} & =2 \times{ }^6 \mathrm{P}_r \\
\frac{10 !}{(10-r+1) !} & =2 \times \frac{6 !}{(6-r) !} \\
\frac{10 !}{6 ! \times 2} & =\frac{(11-r) !}{(6-r) !} \\
\frac{(11-r)(10-r)(9-r)(8-r)(7-r)(6-r) !}{(6-r) !} & =\frac{10 \times 9 \times 8 \times 7 \times 6 !}{6 ! \times 2} \\
\Rightarrow(11-r)(10-r)(9-r)(8-r)(7-r) & =10 \times 9 \times 4 \times 7 \\
& =5 \times 2 \times 3 \times 3 \times 2 \times 2 \times 7 \\
& =7 \times 6 \times 5 \times 4 \times 3 \\
11-r & =7 \\
11-7 & =r \\
r & =4
\end{aligned}
$

Question 3.
(i) Suppose 8 people enter an event in a swimming meet. In how many ways could the gold, silver and bronze prizes be awarded?
Solution:
From 8 persons we have to select and arrange 3 which can be done in ${ }^8 \mathrm{P}_3$ ways So the prizes can be awarded in ${ }^8 \mathrm{P}_3=8 \times 7 \times 6=336$ ways
(ii) Three men have 4 coats, 5 waist coats and 6 caps. In how many ways can they wear them?
Solution:
Selecting and arranging 3 coats from 4 can be done in ${ }^4 \mathrm{P}_3$ ways
Selecting and arranging 3 waist coats from 5 can be done in ${ }^5 \mathrm{P}_3$ ways Selecting and arranging 3 caps from 6 can be done in ${ }^6 \mathrm{P}_3$ ways
$\therefore$ Total number of ways $={ }^4 \mathrm{P}_3 \times{ }^5 \mathrm{P}_3 \times{ }^6 \mathrm{P}_3=172800$ ways
Question 4.
Determine the number of permutations of the letters of the word SIMPLE if all are taken at a time?
Solution:
SIMPLE
Total Number of letters $=6$
They can be arranged in 6 ! ways
$\therefore$ Number of words $=6$ !
$=6 \times 5 \times 4 \times 3 \times 2 \times 1=720$

Question 5.
A test consists of 10 multiple choice questions. In how many ways can the test be answered if
(i) Each question has four choices?
Solution:
Each question has 4 choices. So each questions can be answered in 4 ways.
Number of Questions $=10$
So they can be answered in 410 ways
(ii) The first four questions have three choices and the remaining have five choices?
Solution:
The first four questions have 3 choices. So they can be answered in $3^4$ ways. Remaining 6 questions have 5 choices. So they can be answered in $5^6$ ways.
So all 10 questions can be answered in $3^4 \times 5^6$ ways.
(iii) Question number $\mathrm{n}$ has $\mathrm{n}+1$ choices?
Solution:
Given question $\mathrm{n}$ has $\mathrm{n}+1$ choices
question 1 has $1+1=2$ choices
question 2 has $2+1=3$ choices
question 3 has $3+1=4$ choices
question 4 has $4+1=5$ choices
question 5 has $5+1=6$ choices
question 6 has $6+1=7$ choices
question 7 has $7+1=8$ choices
question 8 has $8+1=9$ choices
question 9 has $9+1=10$ choices
So the number of ways of answering all the 10 questions
$=2 \times 3 \times 4 \times \ldots \times 11=11$ ! ways
Question 6.
A student appears in an objective test which contain 5 multiple choice questions. Each question has four choices out of which one correct answer.
(i) What is the maximum number of different answers can the students give?
Solution:
Selecting a correct answer from the 4 answers can be done in 4 ways.
Total number of questions $=5$ So they can be answered in 45 ways

(ii) How will the answer change if each question may have more than one correct answers?
Solution:
When each question has more than 1 correct answer. Selecting the correct choice from the 4 choice can be done is ${ }^4 C_1$ or ${ }^4 C_2$ or ${ }^4 C_3$ or ${ }^4 C_4$ ways.
$
\begin{gathered}
{ }^4 C_1=4={ }^4 C_3 \\
{ }^4 C_2=\frac{4 \times 3}{2 \times 1}=6 \\
{ }^4 C_4=1 \\
\therefore{ }^4 C_1+{ }^4 C_2+{ }^4 C_3+{ }^4 C_4=4+6+4+1=15
\end{gathered}
$
Each question can be answered in 15 ways.
Number of questions $=5$
$\therefore$ Total number of ways $=15^5$
Question 7.
How many strings can be formed from the letters of the word ARTICLE, so that vowels occupy the even places?
Solution:
ARTICLE
Vowels A, I, $\mathrm{E}=3$
Total number of places $=7$
1234567
Number of even places $=3$
3 Vowels can occupy 3 places in $3 !=3 \times 2 \times 1=6$ ways
Then the remaining 4 letters can be arranged in 4 ! ways
So total number of arrangement $=3 ! \times 4 !=6 \times 24=144$ ways
Question 8.
8 women and 6 men are standing in a line.
(i) How many arrangements are possible if any individual can stand in any position?
Solution:
Total number of persons $=8+6=14$
They can be arranged in 14 ! ways
(ii) In how many arrangements will all 6 men be standing next to one another?
Solution:
There are 6 men and 8 women. To make all 6 men together treat them as 1 unit. Now there are $1+8=9$ persons.

They can be arranged in 9! ways. After this arrangement the 6 men can be arranged in 6 ! ways. So total number of arrangement $=9 ! \times 6$ !
(iii) In how many arrangements will no two men be standing next to one another?
Solution:
Since no two men be together they have to be placed between 8 women and before and after the women.
$\mathrm{w}|\mathrm{w}| \mathrm{w}|\mathrm{w}| \mathrm{w}|\mathrm{w}| \mathrm{w} \mid \mathrm{w}$
There are 9 places so the 6 men can be arranged in the 9 places in ${ }^9 \mathrm{P}_6$ ways.
After this arrangement, the 8 women can be arranged in 8 ! ways.
$\therefore$ Total number of arrangements $=\left({ }^9 \mathrm{P}_6\right) \times 8$ !
Question 9.
Find the distinct permutations of the letters of the word MISSISSIPPI?
Solution:
MISSISSIPPI
Number of letters $=11$
Here $M-1$ time
I - 4 times
$\mathrm{S}-4$ times
$\mathrm{P}-2$ times
So total number of arrangement is of this word $=\frac{11 !}{4 ! 4 ! 2 !}$
$
=\frac{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 !}{4 \times 3 \times 2 \times 1 \times 2 \times 1 \times 4 !}=34650
$

Question 10.
How many ways can the product a2b3c4 be expressed without exponents?
Solution:
$a^2 b^3 c^4=a a b b b c c c$
Number of letters $=9$
$\mathrm{a}=2$ times,
$\mathrm{b}=3$ times,
$\mathrm{c}=4$ times
$\therefore$ Total number of arrangement $=\frac{9 !}{2 ! 3 ! 4 !}=\frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 !}{2 \times 1 \times 3 \times 2 \times 1 \times 4 !}=1260$
Question 11.
In how many ways 4 mathematics books, 3 physics books, 2 chemistry books and 1 biology book can be arranged on a shelf so that all books of the same subjects are together.
Solution:
Number of maths book $=4$
Number of physics books $=3$
Number of chemistry books $=2$
Number of biology books $=1$
Since we want books of the same subjects together, we have to treat all maths books as 1 unit, all physics books as 1 unit, all chemistry books as 1 unit and all biology books as 1 unit. Now total number of units $=4$
They can be arranged in 4 ! ways. After this arrangement.
4 maths book can be arranged in 4 ! ways
3 physics book can be arranged in 3 ! ways
2 chemistry book can be arranged in 2 ! ways and 1 biology book can be arranged in 1 ! way
$\therefore$ Total Number of arrangements $4 ! 4 ! 3 ! 2 !=6912$

Question 12.
In how many ways can the letters of the word SUCCESS be arranged so that all Ss are together?
Solution:
SUCCESS
Number of letters $=7$
Number of ' $S$ ' $=3$
Since we want all ' $S$ ' together treat all $3 \mathrm{~S}$ 's as 1 unit.
Now the remaining letters $=4$
$\therefore$ Total number of unit $=5$
They can be arranged in 5 ! ways of them $\mathrm{C}$ repeats two times.
So total number of arrangements $=\frac{5 !}{2 !}=60$
Question 13.
A coin is tossed 8 times,
(i) How many different sequences of heads and tails are possible?
Solution:
Number of coins tossed $=8$
Number of out come for each toss $=2$
Total number of out comes $=28$
(ii) How many different sequences containing six heads and two tails are possible?
Solution:
Getting 6 heads and 2 tails can be done in ${ }^8 \mathrm{P}_6$ or ${ }^8 \mathrm{P}_2$ ways
$
={ }^8 \mathrm{P}_6={ }^8 \mathrm{P}_2=\frac{8 \times 7}{2 \times 1}=28 \text { ways }
$
Question 14.
How many strings are there using the letters of the word INTERMEDIATE, if

(i) The vowels and consonants are alternative Solution: INTERMEDIATE

The number of ways in which vowels and consonants are alternative $=\frac{6 ! 6 !}{3 ! 2 !}=43200$
(ii) All the vowels are together
Solution:
The number of arrangements:
Keeping all the vowels as a single unit. Now we have $6+1=7$ units which can be arranged in 7 ! ways.
Now the 6 consonants can be arranged in $\frac{6 !}{2 !}$ (T occurs twice) ways in vowels I - repeats thrice and $E$ - repeats twice
So total number of arrangement $=\frac{7 ! 6 !}{3 ! 2 ! 2 !}=151200$
(iii) Vowels are never together (and) (iv) No two vowels are together.
Solution:
Vowels should not be together $=$ No. of all arrangements - No. of all vowels together
Now number of all letters $=\frac{12 !}{3 ! 2 ! 2 !}=19958400$
So number of ways in which No two vowels are together $=19958400-$ Number of ways in which vowels are together $=19958400-151200=19807200$
Question 15.
Each of the digits 1,1,2,3,3 and 4 is written on a separate card. The seven cards are then laid out in a row to form a 6-digit number.
(i) How many distinct 6-digit numbers are there?
Solution:
The given digits are $1,1,2,3,3,4$
The 6 digits can be arranged in 6 ! ways
In which 1 and 3 are repeated twice.
So number of 6 digits numbers $=\frac{6 !}{2 ! 2 !}=\frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 2}=180$

(ii) How many of these 6-digit numbers are even?
Solution:
To find the number even numbers
The digit in unit place is 2 or 4 which can be filled in 2 ways
The remaining digits can be arranged in $\frac{5 !}{2 ! 2 !}$
So number of even numbers $=\frac{5 !}{2 ! 2 !} \times 2=\frac{5 \times 4 \times 3 \times 2 \times 1 \times 2}{2 \times 2}=60$
(iii) How many of these 6-digit numbers are divisible by 4 ?
Solution:
To get a number -f- by 4 the last 2 digits should be -r- by 4 So the last two digits will be 12 or 24 or 32 .
When the last 2 digits are 1 and 2.
The number of 6 digit number $=\frac{4 !}{2 !}=\frac{24}{2}=12$
(remaining digits are $1,3,3,4$ )
When the last 2 digits are 24 the number of 6 digit numbers $=\frac{4 !}{2 ! 2 !}=\frac{24}{4}=6$
When the last 2 digit are 3 and number of 6 digit numbers (remaining number 1, 1,3,4) So there of 6 digit numbers $\div$ by $4=12+6+12=30$
Question 16.
If the letters of the word GARDEN are permuted in all possible ways and the strings thus formed are arranged in the dictionary order, then find the ranks of the words
(i) GARDEN
(ii) DANGER.
Solution:
The given letters are GARDEN.
To find the rank of GARDEN:
The given letters in alphabetical order are A D E G N R

(i)

The rank of GARDEN is 379 To find the rank of DANGER
(ii) The No. of words starting with $\mathrm{A}=5$ ! $=120$

Question 17.
Find the number of strings that can be made using all letters of the word THING. If these words are written as in a dictionary, what will be the $85^{\text {th }}$ string?
Solution:
(i) Number of words formed $=5 !=120$
(ii) The given word is THING
Taking the letters in alphabetical order G H I N T
To find the $85^{\text {th }}$ word

Question 18.
If the letters of the word FUNNY are permuted in all possible ways and the strings thus formed are arranged in the dictionary order, find the rank of the word FUNNY.
Solution:
The given word is FUNNY

Question 19.
Find the sum of all 4-digit numbers that can be formed using digits $1,2,3,4$, and 5 repetitions not allowed?
Solution:
The given digits are $1,2,3,4,5$
The no. of 4 digit numbers

$=5 \times 4 \times 3 \times 2=120$
(i.e) ${ }^5 \mathrm{P}_4=120$
Now we have 120 numbers
So each digit occurs $\frac{120}{5}=24$ times
Sum of the digits $=1+2+3+4+5=15$
Sum of number's in each place $=24 \times 15=360$

Question 20 .
Find the sum of all 4-digit numbers that can be formed using digits $0,2,5,7,8$ without repetition?
Solution:
The given digits are $0,2,5,7,8$
To get the number of 4 digit numbers 1000 's place can be filled in 4 ways (excluding 0 ) 100 's place can be filled in 4 ways (excluding one number and including 0 ) 10 's place can be filled in $(4-1)=3$ ways and unit place can be filled in $(3-1)=2$ ways So the number of 4 digit numbers $=4 \times 4 \times 3 \times 2=96$
To find the sum of 96 numbers:
In 1000's place we have the digits $2,5,7,8$. So each number occurs $\frac{96}{4}=24$ times. Now in 100 's place 0 come 24 times. So the remaining digits $2,5,7,8$ occurs $96-24=\frac{72}{4}$ $=18$ times
Similarly in 10 's place and in unit place 0 occurs 24 times and the remaining digits $2,5,7$, 8 occurs 18 times.
Now sum of the digits $=2+5+7+8=22$
Sum in 1000's place $=22 \times 24=528$
Sum in 100's, 10's and in unit place $=22 \times 18=396$
$\therefore$ Sum of the 4 digit numbers is