1. How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?
3-digit numbers must be formed using 9 digits (1 to 9).
$^{9}P_{3}$ $=\frac{9!}{(9-3)!}$
= $\frac{9!}{6!}$
= $\frac{9\times 8\times 7\times 6!}{6!}$
= $9\times 8\times 7=504$
2. How many 4-digit numbers are there with no digit repeated?
4-digit numbers to be formed using 10 digits (0 to 9), repetition not allowed.
0 cannot be filled in the thousands place. Therefore number of permutations for thousands place is 9.
Remaining 3digits to be filled using remaining 9 digits.
Permutations of such 3-digit number $^{9}P_{3}$ $=\frac{9!}{(9-3)!}$
= $\frac{9!}{6!}$
= $\frac{9\times 8\times 7\times 6!}{6!}$
= $9\times 8\times 7=504$
We have,
number of permutations for thousands place is 9.
number of permutations for remaining 3 digits is 504.
Hence, total number of permutations=9*504=4536
3. How many 3-digit even numbers can be made using the digits 1,2,3,4,6,7, if no digit is repeated?
3-digit number to be formed using 6 digits (1,2,3,4,6,7)
Since we need an even number, units place can be filled in 3 ways.
Remaining 2digits to be filled using remaining 5 digits.
Permutations of such 2-digit number $^{5}P_{2}$ $=\frac{5!}{(5-2)!}$
= $\frac{5!}{3!}$
= $\frac{5\times 4\times 3!}{3!}$
= $5\times 4=20$
We have,
number of permutations for units place is 3.
number of permutations for remaining 2 digits is 20.
Hence, total number of permutations=3*20=60
4. Find the number of 4-digit numbers that can be formed using the digits 1,2,3,4,5 if no digit is repeated. How many of these will be even?
4-digits to be formed using 5 digits (1,2,3,4,5).
number of permutation $^{5}P_{4}$ $=\frac{5!}{(5-4)!}$
= $\frac{5!}{1!}$
= $5\times 4\times 3\times 2\times 1=120$
For even number, units place can be filled with 2,4.
number of permutation $^{2}P_{1}$ $=\frac{2!}{(2-1)!}$
= $\frac{2!}{1!}=2$
Remaining 3digits can be filled with remaining 4 digits.
number of such 3-digit permutation $^{4}P_{3}$ $=\frac{4!}{(4-3)!}$
= $\frac{4!}{1!}$
= $4\times 3\times 2\times 1=24$
Therefore, total number of permutations of even number=2*24=48
5. From a committee of 8 persons, in how many ways can we choose a chairman and a vice chairman assuming one person cannot hold more than one position?
A chairman and a vice chairman to be chosen from 8 person committee.
number of permutation of choosing 2 different persons from 8 persons $^{8}P_{2}$ $=\frac{8!}{(8-2)!}$
= $\frac{8!}{6!}$
= $\frac{8\times 7\times 6!}{6!}$
= $8\times 7=56$
6. Find $n$ if $^{n-1}P_{3}:^{n}P_{4}=1:9$.
$\large \frac{^{n-1}P_{3}}{^{n}P_{4}}=\frac{1}{9}$
$\large \frac{\frac{(n-1)!}{(n-1-3)!}}{\frac{n!}{(n-4)!}}=\frac{1}{9}$
$\large \frac{(n-1)!}{(n-4)!}*\frac{(n-4)!}{n!}=\frac{1}{9}$
$\large \frac{(n-1)!}{n!}=\frac{1}{9}$
$\large \frac{(n-1)!}{n(n-1)!}=\frac{1}{9}$
$ \frac{1}{n}=\frac{1}{9}$
$n=9$
7. Find $r$ if (i) $^{5}P_{r}=2\times ^{6}P_{r-1}$ (ii) $^{5}P_{r}=^{6}P_{r-1}$
(i) $^{5}P_{r}=2\times ^{6}P_{r-1}$
$\large \frac{5!}{(5-r)!}=2\frac{6!}{(6-(r-1))!}$
$\large \frac{5!}{(5-r)!}=2\frac{6\times 5!}{(6-r+1)!}$
$\large \frac{1}{(5-r)!}=2\frac{6}{(7-r)!}$
$\large \frac{1}{(5-r)!}=\frac{12}{(7-r)(6-r)(5-r)!}$
$\large 1=\frac{12}{(7-r)(6-r)}$
$(7-r)(6-r)=12$
$r^{2}-13r+42=12$
$r^{2}-13r+42-12=0$
$r^{2}-13r+30=0$
Simplify,
$(r-10)(r-3)=0$
$r=10$ or $r=3$
$r=10$ is not possible because $r>n$
∴ $r=3$
(ii) $^{5}P_{r}=\times ^{6}P_{r-1}$
$\large \frac{5!}{(5-r)!}=\frac{6!}{(6-(r-1))!}$
$\large \frac{5!}{(5-r)!}=\frac{6\times 5!}{(6-r+1)!}$
$\large \frac{1}{(5-r)!}=\frac{6}{(7-r)!}$
$\large \frac{1}{(5-r)!}=\frac{6}{(7-r)(6-r)(5-r)!}$
$\large 1=\frac{6}{(7-r)(6-r)}$
$(7-r)(6-r)=6$
$r^{2}-13r+42=6$
$r^{2}-13r+42-6=0$
$r^{2}-13r+36=0$
Simplify,
$(r-9)(r-4)=0$
$r=9$ or $r=4$
$r=9$ is not possible because $r>n$
∴ $r=4$
8. How many words, with or without meaning can be formed using all the letters of the word EQUATION, using each letter exactly once?
Number of letters in the word EQUATION is 8.
Number of letters(all different) used is 8.
Number of permutations is $^{8}P_{8}=\frac{8!}{(8-8)!}$
= $\frac{8!}{0!}$
= $8!=40320$
9. How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if (i) 4 letters are used at a time (ii) all letters are used at a time (iii) all letters are used but first letter is a vowel?
Number of letters in the word MONDAY is 6.
(i) 4 letters are used at a time
Number of letters used at a time is 4
Number of permutations is $^{6}P_{4}=\frac{6!}{(6-4)!}$
= $\frac{6!}{2!}$
= $\frac{6\times 5\times 4\times 3\times 2!}{2!}$
= $6\times 5\times 4\times 3$
= $360$
(ii) all letters are used at a time
Number of letters used at a time is 6
Number of permutations is $^{6}P_{6}=\frac{6!}{(6-6)!}$
= $\frac{6!}{0!}$
= $6!=720$
(iii) all letters are used but first letter is a vowel?
Number of letters used at a time is 6
First letter is a vowel and can be filled with O,A i.e. in 2 ways.
Remaining 5 letters are filled with remaining 5 letters $^{5}P_{5}=\frac{5!}{(5-5)!}$
= $\frac{5!}{0!}$
= $5!=120$
Number of permutations is $2*120=240$
10. In how many of the distinct permutations of the letters in MISSISSIPPI do the four I's not come together?
Number of letters in MISSISSIPPI is 11
M=1, I=4, S=4, P=2
Number of permutations is $\frac{11!}{4!4!2!}$
= $\frac{11\times 10\times 9\times 8\times 7\times 6!}{24*24*2}$
= $\frac{11\times 10\times 9\times 8\times 7\times 720}{24*24*2}$
= $34650$
When all I's come together it becomes one object.
So total letters in the word when 4 I's come together is 8.
M=1, I=1, S=4, P=2
Number of permutations is $\frac{8!}{4!2!}$
= $\frac{8\times 7\times 6\times 5\times 4!}{4!2!}$
= $\frac{8\times 7\times 6\times 5}{2}$
= $840$
∴ Number of permutations when four I's not come together is $34650-840=33810$
11. In how many ways can the letters of the word PERMUTATIONS be arranged if the (i) words start with P and end with S (ii) vowels are all together (iii) there are always 4 letters between P and S?
Number of letters in PERMUTATIONS is 12
T=2
(i) words start with P and end with S
P and S are fixed at the extreme ends.
Remaining 10 letters can be arranged in $\frac{10!}{2!}=1814400$
(ii) vowels are all together
Number of vowels is 5 and can be arranged in $5!=120$ ways
when vowels are together it becomes one object.
So total letters in the word when 5 vowels come together is 8.
8 letters can be arranged in $\frac{8!}{2!}=20160$
Number of permutations is $20160*120=2419200$
(iii) there are always 4 letters between P and S
P and S at 1st and 6th place
P and S at 2nd and 7th place
P and S at 3rd and 8th place
P and S at 4th and 9th place
P and S at 5th and 10th place
P and S at 6th and 12th place
Hence, P and S can be put in 7 ways, also P and S can interchange.
∴ Having 4 letters between P and S can be in 7*2=14.
Remaining 10 places to be filled by remaining 10 letters i.e. $\frac{10!}{2!}=1814400$
∴ Number of permutations is $14*1814400=25401600$
3-digit numbers must be formed using 9 digits (1 to 9).
$^{9}P_{3}$ $=\frac{9!}{(9-3)!}$
= $\frac{9!}{6!}$
= $\frac{9\times 8\times 7\times 6!}{6!}$
= $9\times 8\times 7=504$
2. How many 4-digit numbers are there with no digit repeated?
4-digit numbers to be formed using 10 digits (0 to 9), repetition not allowed.
0 cannot be filled in the thousands place. Therefore number of permutations for thousands place is 9.
Remaining 3digits to be filled using remaining 9 digits.
Permutations of such 3-digit number $^{9}P_{3}$ $=\frac{9!}{(9-3)!}$
= $\frac{9!}{6!}$
= $\frac{9\times 8\times 7\times 6!}{6!}$
= $9\times 8\times 7=504$
We have,
number of permutations for thousands place is 9.
number of permutations for remaining 3 digits is 504.
Hence, total number of permutations=9*504=4536
3. How many 3-digit even numbers can be made using the digits 1,2,3,4,6,7, if no digit is repeated?
3-digit number to be formed using 6 digits (1,2,3,4,6,7)
Since we need an even number, units place can be filled in 3 ways.
Remaining 2digits to be filled using remaining 5 digits.
Permutations of such 2-digit number $^{5}P_{2}$ $=\frac{5!}{(5-2)!}$
= $\frac{5!}{3!}$
= $\frac{5\times 4\times 3!}{3!}$
= $5\times 4=20$
We have,
number of permutations for units place is 3.
number of permutations for remaining 2 digits is 20.
Hence, total number of permutations=3*20=60
4. Find the number of 4-digit numbers that can be formed using the digits 1,2,3,4,5 if no digit is repeated. How many of these will be even?
4-digits to be formed using 5 digits (1,2,3,4,5).
number of permutation $^{5}P_{4}$ $=\frac{5!}{(5-4)!}$
= $\frac{5!}{1!}$
= $5\times 4\times 3\times 2\times 1=120$
For even number, units place can be filled with 2,4.
number of permutation $^{2}P_{1}$ $=\frac{2!}{(2-1)!}$
= $\frac{2!}{1!}=2$
Remaining 3digits can be filled with remaining 4 digits.
number of such 3-digit permutation $^{4}P_{3}$ $=\frac{4!}{(4-3)!}$
= $\frac{4!}{1!}$
= $4\times 3\times 2\times 1=24$
Therefore, total number of permutations of even number=2*24=48
5. From a committee of 8 persons, in how many ways can we choose a chairman and a vice chairman assuming one person cannot hold more than one position?
A chairman and a vice chairman to be chosen from 8 person committee.
number of permutation of choosing 2 different persons from 8 persons $^{8}P_{2}$ $=\frac{8!}{(8-2)!}$
= $\frac{8!}{6!}$
= $\frac{8\times 7\times 6!}{6!}$
= $8\times 7=56$
6. Find $n$ if $^{n-1}P_{3}:^{n}P_{4}=1:9$.
$\large \frac{^{n-1}P_{3}}{^{n}P_{4}}=\frac{1}{9}$
$\large \frac{\frac{(n-1)!}{(n-1-3)!}}{\frac{n!}{(n-4)!}}=\frac{1}{9}$
$\large \frac{(n-1)!}{(n-4)!}*\frac{(n-4)!}{n!}=\frac{1}{9}$
$\large \frac{(n-1)!}{n!}=\frac{1}{9}$
$\large \frac{(n-1)!}{n(n-1)!}=\frac{1}{9}$
$ \frac{1}{n}=\frac{1}{9}$
$n=9$
7. Find $r$ if (i) $^{5}P_{r}=2\times ^{6}P_{r-1}$ (ii) $^{5}P_{r}=^{6}P_{r-1}$
(i) $^{5}P_{r}=2\times ^{6}P_{r-1}$
$\large \frac{5!}{(5-r)!}=2\frac{6!}{(6-(r-1))!}$
$\large \frac{5!}{(5-r)!}=2\frac{6\times 5!}{(6-r+1)!}$
$\large \frac{1}{(5-r)!}=2\frac{6}{(7-r)!}$
$\large \frac{1}{(5-r)!}=\frac{12}{(7-r)(6-r)(5-r)!}$
$\large 1=\frac{12}{(7-r)(6-r)}$
$(7-r)(6-r)=12$
$r^{2}-13r+42=12$
$r^{2}-13r+42-12=0$
$r^{2}-13r+30=0$
Simplify,
$(r-10)(r-3)=0$
$r=10$ or $r=3$
$r=10$ is not possible because $r>n$
∴ $r=3$
(ii) $^{5}P_{r}=\times ^{6}P_{r-1}$
$\large \frac{5!}{(5-r)!}=\frac{6!}{(6-(r-1))!}$
$\large \frac{5!}{(5-r)!}=\frac{6\times 5!}{(6-r+1)!}$
$\large \frac{1}{(5-r)!}=\frac{6}{(7-r)!}$
$\large \frac{1}{(5-r)!}=\frac{6}{(7-r)(6-r)(5-r)!}$
$\large 1=\frac{6}{(7-r)(6-r)}$
$(7-r)(6-r)=6$
$r^{2}-13r+42=6$
$r^{2}-13r+42-6=0$
$r^{2}-13r+36=0$
Simplify,
$(r-9)(r-4)=0$
$r=9$ or $r=4$
$r=9$ is not possible because $r>n$
∴ $r=4$
8. How many words, with or without meaning can be formed using all the letters of the word EQUATION, using each letter exactly once?
Number of letters in the word EQUATION is 8.
Number of letters(all different) used is 8.
Number of permutations is $^{8}P_{8}=\frac{8!}{(8-8)!}$
= $\frac{8!}{0!}$
= $8!=40320$
9. How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if (i) 4 letters are used at a time (ii) all letters are used at a time (iii) all letters are used but first letter is a vowel?
Number of letters in the word MONDAY is 6.
(i) 4 letters are used at a time
Number of letters used at a time is 4
Number of permutations is $^{6}P_{4}=\frac{6!}{(6-4)!}$
= $\frac{6!}{2!}$
= $\frac{6\times 5\times 4\times 3\times 2!}{2!}$
= $6\times 5\times 4\times 3$
= $360$
(ii) all letters are used at a time
Number of letters used at a time is 6
Number of permutations is $^{6}P_{6}=\frac{6!}{(6-6)!}$
= $\frac{6!}{0!}$
= $6!=720$
(iii) all letters are used but first letter is a vowel?
Number of letters used at a time is 6
First letter is a vowel and can be filled with O,A i.e. in 2 ways.
Remaining 5 letters are filled with remaining 5 letters $^{5}P_{5}=\frac{5!}{(5-5)!}$
= $\frac{5!}{0!}$
= $5!=120$
Number of permutations is $2*120=240$
10. In how many of the distinct permutations of the letters in MISSISSIPPI do the four I's not come together?
Number of letters in MISSISSIPPI is 11
M=1, I=4, S=4, P=2
Number of permutations is $\frac{11!}{4!4!2!}$
= $\frac{11\times 10\times 9\times 8\times 7\times 6!}{24*24*2}$
= $\frac{11\times 10\times 9\times 8\times 7\times 720}{24*24*2}$
= $34650$
When all I's come together it becomes one object.
So total letters in the word when 4 I's come together is 8.
M=1, I=1, S=4, P=2
Number of permutations is $\frac{8!}{4!2!}$
= $\frac{8\times 7\times 6\times 5\times 4!}{4!2!}$
= $\frac{8\times 7\times 6\times 5}{2}$
= $840$
∴ Number of permutations when four I's not come together is $34650-840=33810$
11. In how many ways can the letters of the word PERMUTATIONS be arranged if the (i) words start with P and end with S (ii) vowels are all together (iii) there are always 4 letters between P and S?
Number of letters in PERMUTATIONS is 12
T=2
(i) words start with P and end with S
P and S are fixed at the extreme ends.
Remaining 10 letters can be arranged in $\frac{10!}{2!}=1814400$
(ii) vowels are all together
Number of vowels is 5 and can be arranged in $5!=120$ ways
when vowels are together it becomes one object.
So total letters in the word when 5 vowels come together is 8.
8 letters can be arranged in $\frac{8!}{2!}=20160$
Number of permutations is $20160*120=2419200$
(iii) there are always 4 letters between P and S
P and S at 1st and 6th place
P and S at 2nd and 7th place
P and S at 3rd and 8th place
P and S at 4th and 9th place
P and S at 5th and 10th place
P and S at 6th and 12th place
Hence, P and S can be put in 7 ways, also P and S can interchange.
∴ Having 4 letters between P and S can be in 7*2=14.
Remaining 10 places to be filled by remaining 10 letters i.e. $\frac{10!}{2!}=1814400$
∴ Number of permutations is $14*1814400=25401600$
0 Comments