1. How many words, with or without meaning, each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER ?
5 letter word to be formed from 8 letters of word Daughter.
2 vowels to be selected from 3 vowels (A,U,E), $^{3}C_{2}$
3 consonants to be selected from 5 consonants (D,G,H,T,R), $^{5}C_{3}$
Therefore number of ways of selection $^{3}C_{2}$*$^{5}C_{3}$
= $\frac{3!}{2!(3-2)!}*\frac{5!}{3!(5-3)!}$
= $\frac{3!}{2!1!}*\frac{5!}{3!2!}$
= $\frac{3\times 2!}{2!}*\frac{5\times 4\times 3!}{3!2!}$
= $3*\frac{5\times 4}{2!}$
= $3*5*2$
= $30$
Each 5 letter word can be arranged in 5! ways.
∴ Total number of words=5!*30=3600
2. How many words, with or without meaning, can be formed using all the letters of the word EQUATION at a time so that the vowels and consonants occur together?
8 letter word to be formed from 8 letters of word EQUATION
Vowels occurring together is one object
Consonants occurring together is one object
vowels and consonants occurring together can be arranged in 2! ways
5 vowels can be arranged in 5! ways
3 consonants can be arranged in 3! ways
∴ number of words = 2!*5!*3!=1440
3. A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of: (i) exactly 3 girls ? (ii) at least 3 girls ? (iii) at most 3 girls ?
(i) exactly 3 girls
3 girls to be selected from 4 girls $^{4}C_{3}$
4 boys to be selected from 9 boys $^{9}C_{4}$
$^{4}C_{3}$*$^{9}C_{4}$
= $\frac{4!}{3!(4-3)!}*\frac{9!}{4!(9-4)!}$
= $\frac{4!}{3!1!}*\frac{9!}{4!5!}$
= $\frac{4\times 3!}{3!}*\frac{9\times 8\times 7\times 6\times 5!}{4!5!}$
= $4*\frac{9\times 8\times 7\times 6}{4!}$
= $4*126$
= $504$
(ii) atleast 3 girls
either 3 girls or 4 girls
3 girls to be selected from 4 girls $^{4}C_{3}$
4 boys to be selected from 9 boys $^{9}C_{4}$
$^{4}C_{3}$*$^{9}C_{4}$
= $\frac{4!}{3!(4-3)!}*\frac{9!}{4!(9-4)!}$
= $\frac{4!}{3!1!}*\frac{9!}{4!5!}$
= $\frac{4\times 3!}{3!}*\frac{9\times 8\times 7\times 6\times 5!}{4!5!}$
= $4*\frac{9\times 8\times 7\times 6}{4!}$
= $4*126$
= $504$
OR
4 girls to be selected from 4 girls $^{4}C_{4}$
3 boys to be selected from 9 boys $^{9}C_{3}$
$^{4}C_{4}$*$^{9}C_{3}$
= $\frac{4!}{4!(4-4)!}*\frac{9!}{3!(9-3)!}$
= $1*\frac{9!}{3!6!}$
= $\frac{9\times 8\times 7\times 6!}{3!6!}$
= $\frac{9\times 8\times 7}{3!}$
= $84$
Number of ways of selection is 504+84=588
(iii) atmost 3 girls
either 0,1,2 or 3
0 girls to be selected from 4 girls $^{4}C_{0}$
7 boys to be selected from 9 boys $^{9}C_{7}$
$^{4}C_{0}$*$^{9}C_{7}$
= $\frac{4!}{0!(4-0)!}*\frac{9!}{7!(9-7)!}$
= $\frac{4!}{4!}*\frac{9!}{7!2!}$
= $1*\frac{9\times 8\times 7!}{7!2!}$
= $\frac{9\times 8}{2!}$
= $36$
OR
1 girls to be selected from 4 girls $^{4}C_{1}$
6 boys to be selected from 9 boys $^{9}C_{6}$
$^{4}C_{1}$*$^{9}C_{6}$
= $\frac{4!}{1!(4-1)!}*\frac{9!}{6!(9-6)!}$
= $\frac{4!}{1!3!}*\frac{9!}{6!3!}$
= $\frac{4\times 3!}{3!}*\frac{9\times 8\times 7\times 6!}{6!3!}$
= $4*\frac{9\times 8\times 7}{3!}$
= $336$
OR
2 girls to be selected from 4 girls $^{4}C_{2}$
5 boys to be selected from 9 boys $^{9}C_{5}$
$^{4}C_{2}$*$^{9}C_{5}$
= $\frac{4!}{2!(4-2)!}*\frac{9!}{5!(9-5)!}$
= $\frac{4!}{2!2!}*\frac{9!}{5!4!}$
= $\frac{4\times 3\times 2!}{2!2!}*\frac{9\times 8\times 7\times 6\times 5!}{4!5!}$
= $\frac{4\times 3}{2!}*\frac{9\times 8\times 7\times 6}{4!}$
= $6*126$
= $756$
OR
3 girls to be selected from 4 girls $^{4}C_{3}$
4 boys to be selected from 9 boys $^{9}C_{4}$
$^{4}C_{3}$*$^{9}C_{4}$
= $\frac{4!}{3!(4-3)!}*\frac{9!}{4!(9-4)!}$
= $\frac{4!}{3!1!}*\frac{9!}{4!5!}$
= $\frac{4\times 3!}{3!}*\frac{9\times 8\times 7\times 6\times 5!}{4!5!}$
= $4*\frac{9\times 8\times 7\times 6}{4!}$
= $4*126$
= $504$
Number of ways of selection = 36+336+756+504=1632
4. If the different permutations of all the letter of the word EXAMINATION are listed as in a dictionary, how many words are there in this list before the first word starting with E ?
In the given word EXAMINATION, total letter=11
A=2, I=2, N=2
Words before starting with E in dictionary will be words starting with A.
A fixed at extreme left position.
Number of ways of arrangement of remaining 10 letters = $\frac{10!}{2!2!}$
= $907200$
5. How many 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7 and 9 which are divisible by 10 and no digit is repeated ?
6-digit number to be and must be divisible by 10.
Units digit must be 0 to be divisible by 10.
Remaining 5 places to be filled from remaining 5 digits in $^{5}P_{5}$ ways
$^{5}P_{5}$
= $\frac{5!}{(5-5)!}$
= $\frac{5!}{0!}
= $5!=120$
6. The English alphabet has 5 vowels and 21 consonants. How many words with two different vowels and 2 different consonants can be formed from the alphabet ?
2 vowels to be selected from 5 vowels, $^{5}C_{2}$
2 consonants to be selected from 21 consonants, $^{21}C_{2}$
These 4 letters can be arranged in 4! ways.
$^{5}C_{2}$*$^{21}C_{2}$
= $\frac{5!}{2!(5-2)!}*\frac{21!}{2!(21-2)!}$
= $\frac{5!}{2!3!}*\frac{21!}{2!19!}$
= $\frac{5\times 4\times 3!}{2!3!}*\frac{21\times 20\times 19!}{2!19!}$
= $\frac{5\times 4}{2!}*\frac{21\times 20}{2!}$
= $10*210$
= $2100$
Number of combinations = 2100*4!= 50400
7. In an examination, a question paper consists of 12 questions divided into two parts i.e., Part I and Part II, containing 5 and 7 questions, respectively. A student is required to attempt 8 questions in all, selecting at least 3 from each part. In how many ways can a student select the questions ?
8 questions to be selected from 12 questions.
at least 3 questions from each section.
Therefore,
3 from part I out of 5 questions $^{5}C_{3}$
5 from part II out of 7 questions $^{7}C_{5}$
$^{5}C_{3}*^{7}C_{5}$
= $\frac{5!}{3!(5-3)!}*\frac{7!}{5!(7-5)!}$
= $\frac{5!}{3!2!}*\frac{7!}{5!2!}$
= $\frac{5\times 4\times 3!}{3!2!}*\frac{7\times 6\times 5!}{5!2!}$
= $\frac{5\times 4}{2!}*\frac{7\times 6}{2!}$
= $10*21$
= $210$
OR
4 from part I out of 5 questions $^{5}C_{4}$
4 from part II out of 7 questions $^{7}C_{4}$
$^{5}C_{4}*^{7}C_{4}$
= $\frac{5!}{4!(5-4)!}*\frac{7!}{4!(7-4)!}$
= $\frac{5!}{4!1!}*\frac{7!}{4!3!}$
= $\frac{5\times 4!}{4!}*\frac{7\times 6\times 5\times 4!}{4!3!}$
= $5*\frac{7\times 6\times 5}{3!}$
= $5*35$
= $175$
OR
5 from part I out of 5 questions $^{5}C_{5}$
3 from part II out of 7 questions $^{7}C_{3}$
$^{5}C_{5}*^{7}C_{3}$
= $1*\frac{7!}{3!(7-3)!}$
= $\frac{7!}{3!4!}$
= $\frac{7\times 6\times 5\times 4!}{3!4!}$
= $\frac{7\times 6\times 5}{3!}$
= $35$
Number of ways of selection = 210+175+35= 420
8. Determine the number of 5-card combinations out of a deck of 52 cards if each selection of 5 cards has exactly one king.
5 cards to be selected from 52 cards
1 card must be King.
Therefore,
1 king out of 4 kings, $^{4}C_{1}$
remaining 4 cards out of 48 cards, $^{48}C_{4}$
$^{4}C_{1}*^{48}C_{4}$
= $\frac{4!}{1!(4-1)!}*\frac{48!}{4!(48-4)!}$
= $\frac{4!}{1!3!}*\frac{48!}{4!44!}$
= $\frac{4\times 3!}{3!}*\frac{48\times 47\times 46\times 45\times 44!}{4!44!}$
= $4*\frac{48\times 47\times 46\times 45}{4!}$
= $4*194580$
= $778320$
9. It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible ?
9 seats to be filled by 5 men and 4 women
1 2 3 4 5 6 7 8 9
MWMWMWMWM
4 women can be arranged in 4! ways
4 men can be arranged in 5! ways.
Numbre of arrangemnts = 4!*5! = 2880
10. From a class of 25 students, 10 are to be chosen for an excursion party. There are 3 students who decide that either all of them will join or none of them will join. In how many ways can the excursion party be chosen ?
10 students to be chosen from 25 students.
3 students decide whether they join or not.
(i) 3 students will join
$^{3}C_{3}*^{22}C_{7}$
= $1*\frac{22!}{7!(22-7)!}$
= $\frac{22!}{7!15!}$
= $\frac{22\times 21\times 20\times 19\times 18\times 17\times 16\times 15!}{7!15!}$
= $\frac{22\times 21\times 20\times 19\times 18\times 17\times 16}{7!}$
= $\frac{22\times 21\times 20\times 19\times 18\times 17\times 16}{7\times 6\times 5\times 4\times 3\times 2}$
= $170544$
(ii) 3 students will not join
$^{3}C_{0}*^{22}C_{10}$
= $1*\frac{22!}{10!(22-10)!}$
= $\frac{22!}{10!12!}$
= $\frac{22\times 21\times 20\times 19\times 18\times 17\times 16\times 15\times 14\times 13\times 12!}{10!12!}$
= $\frac{22\times 21\times 20\times 19\times 18\times 17\times 16\times 15\times 14\times 13}{10!}$
= $\frac{22\times 21\times 20\times 19\times 18\times 17\times 16\times 15\times 14\times 13}{10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2}$
= $646646$
Number of ways = 170544+646646=817190
11. In how many ways can the letters of the word ASSASSINATION be arranged so that all the S’s are together ?
In the word ASSASSINATION, total letters=13,
A=3, S=4, I=2, N=2
4 S taken as one object
S object and remaining 9 objects will account for 10 objects.
number of arrangement = $\frac{10!}{3!2!2!}$
= $\frac{10\times 9\times 8\times 7\times 6\times 5!}{6*2*2}$
= $151200$
5 letter word to be formed from 8 letters of word Daughter.
2 vowels to be selected from 3 vowels (A,U,E), $^{3}C_{2}$
3 consonants to be selected from 5 consonants (D,G,H,T,R), $^{5}C_{3}$
Therefore number of ways of selection $^{3}C_{2}$*$^{5}C_{3}$
= $\frac{3!}{2!(3-2)!}*\frac{5!}{3!(5-3)!}$
= $\frac{3!}{2!1!}*\frac{5!}{3!2!}$
= $\frac{3\times 2!}{2!}*\frac{5\times 4\times 3!}{3!2!}$
= $3*\frac{5\times 4}{2!}$
= $3*5*2$
= $30$
Each 5 letter word can be arranged in 5! ways.
∴ Total number of words=5!*30=3600
2. How many words, with or without meaning, can be formed using all the letters of the word EQUATION at a time so that the vowels and consonants occur together?
8 letter word to be formed from 8 letters of word EQUATION
Vowels occurring together is one object
Consonants occurring together is one object
vowels and consonants occurring together can be arranged in 2! ways
5 vowels can be arranged in 5! ways
3 consonants can be arranged in 3! ways
∴ number of words = 2!*5!*3!=1440
3. A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of: (i) exactly 3 girls ? (ii) at least 3 girls ? (iii) at most 3 girls ?
(i) exactly 3 girls
3 girls to be selected from 4 girls $^{4}C_{3}$
4 boys to be selected from 9 boys $^{9}C_{4}$
$^{4}C_{3}$*$^{9}C_{4}$
= $\frac{4!}{3!(4-3)!}*\frac{9!}{4!(9-4)!}$
= $\frac{4!}{3!1!}*\frac{9!}{4!5!}$
= $\frac{4\times 3!}{3!}*\frac{9\times 8\times 7\times 6\times 5!}{4!5!}$
= $4*\frac{9\times 8\times 7\times 6}{4!}$
= $4*126$
= $504$
(ii) atleast 3 girls
either 3 girls or 4 girls
3 girls to be selected from 4 girls $^{4}C_{3}$
4 boys to be selected from 9 boys $^{9}C_{4}$
$^{4}C_{3}$*$^{9}C_{4}$
= $\frac{4!}{3!(4-3)!}*\frac{9!}{4!(9-4)!}$
= $\frac{4!}{3!1!}*\frac{9!}{4!5!}$
= $\frac{4\times 3!}{3!}*\frac{9\times 8\times 7\times 6\times 5!}{4!5!}$
= $4*\frac{9\times 8\times 7\times 6}{4!}$
= $4*126$
= $504$
OR
4 girls to be selected from 4 girls $^{4}C_{4}$
3 boys to be selected from 9 boys $^{9}C_{3}$
$^{4}C_{4}$*$^{9}C_{3}$
= $\frac{4!}{4!(4-4)!}*\frac{9!}{3!(9-3)!}$
= $1*\frac{9!}{3!6!}$
= $\frac{9\times 8\times 7\times 6!}{3!6!}$
= $\frac{9\times 8\times 7}{3!}$
= $84$
Number of ways of selection is 504+84=588
(iii) atmost 3 girls
either 0,1,2 or 3
0 girls to be selected from 4 girls $^{4}C_{0}$
7 boys to be selected from 9 boys $^{9}C_{7}$
$^{4}C_{0}$*$^{9}C_{7}$
= $\frac{4!}{0!(4-0)!}*\frac{9!}{7!(9-7)!}$
= $\frac{4!}{4!}*\frac{9!}{7!2!}$
= $1*\frac{9\times 8\times 7!}{7!2!}$
= $\frac{9\times 8}{2!}$
= $36$
OR
1 girls to be selected from 4 girls $^{4}C_{1}$
6 boys to be selected from 9 boys $^{9}C_{6}$
$^{4}C_{1}$*$^{9}C_{6}$
= $\frac{4!}{1!(4-1)!}*\frac{9!}{6!(9-6)!}$
= $\frac{4!}{1!3!}*\frac{9!}{6!3!}$
= $\frac{4\times 3!}{3!}*\frac{9\times 8\times 7\times 6!}{6!3!}$
= $4*\frac{9\times 8\times 7}{3!}$
= $336$
OR
2 girls to be selected from 4 girls $^{4}C_{2}$
5 boys to be selected from 9 boys $^{9}C_{5}$
$^{4}C_{2}$*$^{9}C_{5}$
= $\frac{4!}{2!(4-2)!}*\frac{9!}{5!(9-5)!}$
= $\frac{4!}{2!2!}*\frac{9!}{5!4!}$
= $\frac{4\times 3\times 2!}{2!2!}*\frac{9\times 8\times 7\times 6\times 5!}{4!5!}$
= $\frac{4\times 3}{2!}*\frac{9\times 8\times 7\times 6}{4!}$
= $6*126$
= $756$
OR
3 girls to be selected from 4 girls $^{4}C_{3}$
4 boys to be selected from 9 boys $^{9}C_{4}$
$^{4}C_{3}$*$^{9}C_{4}$
= $\frac{4!}{3!(4-3)!}*\frac{9!}{4!(9-4)!}$
= $\frac{4!}{3!1!}*\frac{9!}{4!5!}$
= $\frac{4\times 3!}{3!}*\frac{9\times 8\times 7\times 6\times 5!}{4!5!}$
= $4*\frac{9\times 8\times 7\times 6}{4!}$
= $4*126$
= $504$
Number of ways of selection = 36+336+756+504=1632
4. If the different permutations of all the letter of the word EXAMINATION are listed as in a dictionary, how many words are there in this list before the first word starting with E ?
In the given word EXAMINATION, total letter=11
A=2, I=2, N=2
Words before starting with E in dictionary will be words starting with A.
A fixed at extreme left position.
Number of ways of arrangement of remaining 10 letters = $\frac{10!}{2!2!}$
= $907200$
5. How many 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7 and 9 which are divisible by 10 and no digit is repeated ?
6-digit number to be and must be divisible by 10.
Units digit must be 0 to be divisible by 10.
Remaining 5 places to be filled from remaining 5 digits in $^{5}P_{5}$ ways
$^{5}P_{5}$
= $\frac{5!}{(5-5)!}$
= $\frac{5!}{0!}
= $5!=120$
6. The English alphabet has 5 vowels and 21 consonants. How many words with two different vowels and 2 different consonants can be formed from the alphabet ?
2 vowels to be selected from 5 vowels, $^{5}C_{2}$
2 consonants to be selected from 21 consonants, $^{21}C_{2}$
These 4 letters can be arranged in 4! ways.
$^{5}C_{2}$*$^{21}C_{2}$
= $\frac{5!}{2!(5-2)!}*\frac{21!}{2!(21-2)!}$
= $\frac{5!}{2!3!}*\frac{21!}{2!19!}$
= $\frac{5\times 4\times 3!}{2!3!}*\frac{21\times 20\times 19!}{2!19!}$
= $\frac{5\times 4}{2!}*\frac{21\times 20}{2!}$
= $10*210$
= $2100$
Number of combinations = 2100*4!= 50400
7. In an examination, a question paper consists of 12 questions divided into two parts i.e., Part I and Part II, containing 5 and 7 questions, respectively. A student is required to attempt 8 questions in all, selecting at least 3 from each part. In how many ways can a student select the questions ?
8 questions to be selected from 12 questions.
at least 3 questions from each section.
Therefore,
3 from part I out of 5 questions $^{5}C_{3}$
5 from part II out of 7 questions $^{7}C_{5}$
$^{5}C_{3}*^{7}C_{5}$
= $\frac{5!}{3!(5-3)!}*\frac{7!}{5!(7-5)!}$
= $\frac{5!}{3!2!}*\frac{7!}{5!2!}$
= $\frac{5\times 4\times 3!}{3!2!}*\frac{7\times 6\times 5!}{5!2!}$
= $\frac{5\times 4}{2!}*\frac{7\times 6}{2!}$
= $10*21$
= $210$
OR
4 from part I out of 5 questions $^{5}C_{4}$
4 from part II out of 7 questions $^{7}C_{4}$
$^{5}C_{4}*^{7}C_{4}$
= $\frac{5!}{4!(5-4)!}*\frac{7!}{4!(7-4)!}$
= $\frac{5!}{4!1!}*\frac{7!}{4!3!}$
= $\frac{5\times 4!}{4!}*\frac{7\times 6\times 5\times 4!}{4!3!}$
= $5*\frac{7\times 6\times 5}{3!}$
= $5*35$
= $175$
OR
5 from part I out of 5 questions $^{5}C_{5}$
3 from part II out of 7 questions $^{7}C_{3}$
$^{5}C_{5}*^{7}C_{3}$
= $1*\frac{7!}{3!(7-3)!}$
= $\frac{7!}{3!4!}$
= $\frac{7\times 6\times 5\times 4!}{3!4!}$
= $\frac{7\times 6\times 5}{3!}$
= $35$
Number of ways of selection = 210+175+35= 420
8. Determine the number of 5-card combinations out of a deck of 52 cards if each selection of 5 cards has exactly one king.
5 cards to be selected from 52 cards
1 card must be King.
Therefore,
1 king out of 4 kings, $^{4}C_{1}$
remaining 4 cards out of 48 cards, $^{48}C_{4}$
$^{4}C_{1}*^{48}C_{4}$
= $\frac{4!}{1!(4-1)!}*\frac{48!}{4!(48-4)!}$
= $\frac{4!}{1!3!}*\frac{48!}{4!44!}$
= $\frac{4\times 3!}{3!}*\frac{48\times 47\times 46\times 45\times 44!}{4!44!}$
= $4*\frac{48\times 47\times 46\times 45}{4!}$
= $4*194580$
= $778320$
9. It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible ?
9 seats to be filled by 5 men and 4 women
1 2 3 4 5 6 7 8 9
MWMWMWMWM
4 women can be arranged in 4! ways
4 men can be arranged in 5! ways.
Numbre of arrangemnts = 4!*5! = 2880
10. From a class of 25 students, 10 are to be chosen for an excursion party. There are 3 students who decide that either all of them will join or none of them will join. In how many ways can the excursion party be chosen ?
10 students to be chosen from 25 students.
3 students decide whether they join or not.
(i) 3 students will join
$^{3}C_{3}*^{22}C_{7}$
= $1*\frac{22!}{7!(22-7)!}$
= $\frac{22!}{7!15!}$
= $\frac{22\times 21\times 20\times 19\times 18\times 17\times 16\times 15!}{7!15!}$
= $\frac{22\times 21\times 20\times 19\times 18\times 17\times 16}{7!}$
= $\frac{22\times 21\times 20\times 19\times 18\times 17\times 16}{7\times 6\times 5\times 4\times 3\times 2}$
= $170544$
(ii) 3 students will not join
$^{3}C_{0}*^{22}C_{10}$
= $1*\frac{22!}{10!(22-10)!}$
= $\frac{22!}{10!12!}$
= $\frac{22\times 21\times 20\times 19\times 18\times 17\times 16\times 15\times 14\times 13\times 12!}{10!12!}$
= $\frac{22\times 21\times 20\times 19\times 18\times 17\times 16\times 15\times 14\times 13}{10!}$
= $\frac{22\times 21\times 20\times 19\times 18\times 17\times 16\times 15\times 14\times 13}{10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2}$
= $646646$
Number of ways = 170544+646646=817190
11. In how many ways can the letters of the word ASSASSINATION be arranged so that all the S’s are together ?
In the word ASSASSINATION, total letters=13,
A=3, S=4, I=2, N=2
4 S taken as one object
S object and remaining 9 objects will account for 10 objects.
number of arrangement = $\frac{10!}{3!2!2!}$
= $\frac{10\times 9\times 8\times 7\times 6\times 5!}{6*2*2}$
= $151200$
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