Exercise 4.1-Additional Questions - Chapter 4 Combinatorics and Mathematical Induction 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Additional Questions Solved
Question 1.
If the letter of the word 'RACHIT' are arranged in all possible ways as listed in dictionary, then what is the rank of the word 'RACHIT'?
Solution:
The alphabetical order of RACHIT is A, C, H, I, R and T
Number of words beginning with $\mathrm{A}=5$ !
Number of words beginning with $\mathrm{C}=5$ !
Number of words beginning with $\mathrm{H}=5$ !
Number of words beginning with $1=5$ !
and Number of words beginning with $\mathrm{R}$ (i.e) $\mathrm{RACHIT}=1$
$\therefore$ The rank of the word 'RACHIT' in the dictionary $=5 !+5 !+5 !+5 !+1=4 \times 5 !+1$
$=4 \times 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1+1=4 \times 120+1=480+1=481$
Question 2.
Find the number of positive integers greater than 6000 and less than 7000 which are divisible by 5 , provided that no digit is to be repeated.
Solution:

Any number divisible by 5 , its unit place must have 0 or 5 . We have to find 4-digit number greater than 6000 and less than 7000 .
So, the unit place can be filled with 2 ways ( 0 or 5 ) since, repetition is not allowed.
$\therefore$ Tens place can be filled with 7 ways and hundreds place can be filled with 8 ways.
But the required number is greater than 6000 and less than 7000 . So, thousand place can be filled with 1 digits (i.e) 6.

So, the total number of integers $=1 \times 8 \times 7 \times 2=112$
Hence, the required number of integers $=112$
Question 3.
Find the number of integers greater than 7000 that can be formed with the digits $3,5,7,8$ and 9 where no digits are repeated.
Solution:
Given that all the 5 digit numbers are greater than 7000 .
So, the ways of forming 5 -digit numbers $=5 \times 4 \times 3 \times 2 \times 1=120$
Now, all the four digit number greater than 7000 can be formed as follows.
Thousand place can be filled with 3 ways
Hundred place can be filled with 4 ways
Tenths place can be filled with 3 ways
Units place can be filled with 2 ways
So, the total number of 4-digits numbers $=3 \times 4 \times 3 \times 2=72$
$\therefore$ Total number of integers $=120+72=192$
Hence, the required number of integers $=192$
Question 4.
How many words (with or without dictionary meaning) can be made from the letters of the word MONDAY, assuming that no letter is repeated, if

Solution:
(a) 4 letters are used at a time $={ }^6 \mathrm{P}_4=\frac{6 !}{2 !}=360$
(b) All letters are used at a time $={ }^6 \mathrm{P}_6=6 !=720$
(c) All letters are used but the first is a vowel $=2 \times 5$ ! $=2 \times 120=240$ Hence, the required matching is
(a) $\leftrightarrow$ (iii), (b) $\leftrightarrow$ (i), (c) $\leftrightarrow$ (ii)
Question 5.
Five boys and 5 girls form a line. Find the number of ways of making the seating arrangement under the following condition.

Solution:
(a) Total number of arrangement when boys and girls alternate : $=(5 !)^2+(5 !)^2$
(b) No two girls sit together $=5$ ! 6 !
(c) All the girls sit together $=2 ! 5 ! 5$ !
(d) All the girls sit never together $=10 !-5$ ! 6 !
Hence, the required matching is (a) $\leftrightarrow$ (iii), (b) $\leftrightarrow$ (i), (c) $\leftrightarrow$ (iv), (d) $\leftrightarrow$ (ii)
Question 6.
How many automobile license plates can be made, if each plate contains two different letters followed by three different digits?
Solution:
We have 26 English alphabet and 10 digits (0 to 9)
Since, it is given that each plate contains 2 different letters followed by 3 different digits.
$\therefore$ Number of arrangement of 26 letter taken 2 at a time
$
={ }^{26} \mathrm{P}_2=\frac{26 !}{(26-2) !}=\frac{26 !}{24 !}=\frac{26 \cdot 25.24 !}{24 !}=650
$
Three digit number can be formed out of 10 digit $={ }^{10} \mathrm{P}_3$
$
=\frac{10 !}{7 !}=\frac{10 \cdot 9 \cdot 8 \cdot 7 !}{7 !}=720
$
Total number of license plates $=650 \times 720=468000$
Hence, the required number of plates $=468000$.