Q1
How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that
(i) repetition of the digits is allowed?
(ii) repetition of the digits is not allowed?
(i) repetition of the digits is allowed?
(ii) repetition of the digits is not allowed?
Answer:
(i) When repetition is allowed:
We need to form a 3-digit number. There are 3 places: unit's, ten's, and hundred's.
- Hundred's place can be filled in 5 ways.
- Ten's place can be filled in 5 ways.
- Unit's place can be filled in 5 ways.
Total numbers = \( 5 \times 5 \times 5 = 125 \).
(ii) When repetition is not allowed:
- Hundred's place can be filled in 5 ways.
- Ten's place can be filled in 4 ways (one used).
- Unit's place can be filled in 3 ways (two used).
Total numbers = \( 5 \times 4 \times 3 = 60 \).
Q2
How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?
Answer:
For a number to be even, the unit's digit must be 2, 4, or 6.
- Unit's place: 3 options (2, 4, 6).
- Ten's place: 6 options (since repetition is allowed).
- Hundred's place: 6 options.
Total numbers = \( 6 \times 6 \times 3 = 108 \).
Q3
How many 4-letter code can be formed using the first 10 letters of the English alphabet, if no letter can be repeated?
Answer:
We need to fill 4 places using 10 distinct letters.
- 1st place: 10 ways
- 2nd place: 9 ways
- 3rd place: 8 ways
- 4th place: 7 ways
Total codes = \( 10 \times 9 \times 8 \times 7 = 5040 \).
Q4
How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once?
Answer:
The first two digits are fixed as 6 and 7.
We need to fill the remaining 3 places using the remaining 8 digits (0, 1, 2, 3, 4, 5, 8, 9).
- 3rd place: 8 ways
- 4th place: 7 ways
- 5th place: 6 ways
Total numbers = \( 1 \times 1 \times 8 \times 7 \times 6 = 336 \).
Q5
A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?
Answer:
Each toss has 2 possible outcomes (Head or Tail).
Total outcomes for 3 tosses = \( 2 \times 2 \times 2 = 8 \).
Q6
Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?
Answer:
We need to fill 2 positions (upper and lower) with 5 available flags.
- Upper flag: 5 options.
- Lower flag: 4 options (since flags must be different).
Total signals = \( 5 \times 4 = 20 \).
Q1
Evaluate (i) 8! (ii) 4! – 3!
Answer:
(i) \( 8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40320 \).
(ii) \( 4! - 3! = (4 \times 3 \times 2 \times 1) - (3 \times 2 \times 1) = 24 - 6 = 18 \).
Q2
Is 3! + 4! = 7! ?
Answer:
LHS: \( 3! + 4! = 6 + 24 = 30 \).
RHS: \( 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040 \).
Since \( 30 \ne 5040 \), No, \( 3! + 4! \ne 7! \).
Q3
Compute \( \frac{8!}{6! \times 2!} \)
Answer:
\( \frac{8!}{6! \times 2!} = \frac{8 \times 7 \times 6!}{6! \times (2 \times 1)} = \frac{56}{2} = 28 \).
Q4
If \( \frac{1}{6!} + \frac{1}{7!} = \frac{x}{8!} \), find \( x \).
Answer:
\( \frac{1}{6!} + \frac{1}{7 \times 6!} = \frac{x}{8 \times 7 \times 6!} \)
Multiplying both sides by \( 8! \):
\( \frac{8!}{6!} + \frac{8!}{7!} = x \)
\( (8 \times 7) + 8 = x \)
\( 56 + 8 = x \implies x = 64 \).
Q5
Evaluate \( \frac{n!}{(n-r)!} \), when (i) \( n = 6, r = 2 \) (ii) \( n = 9, r = 5 \).
Answer:
(i) \( \frac{6!}{(6-2)!} = \frac{6!}{4!} = \frac{6 \times 5 \times 4!}{4!} = 30 \).
(ii) \( \frac{9!}{(9-5)!} = \frac{9!}{4!} = 9 \times 8 \times 7 \times 6 \times 5 = 15120 \).
Q1
How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?
Answer:
Number of digits \( n = 9 \). Places to fill \( r = 3 \).
Permutations \( ^9P_3 = \frac{9!}{(9-3)!} = \frac{9!}{6!} = 9 \times 8 \times 7 = 504 \).
Q2
How many 4-digit numbers are there with no digit repeated?
Answer:
Total digits available: 0, 1, ..., 9 (10 digits).
We need to fill 4 places: Thousands (T), Hundreds (H), Tens (T), Units (U).
- Thousands place: 9 options (1-9, cannot be 0).
- Hundreds place: 9 options (0 and remaining 8 digits).
- Tens place: 8 options.
- Units place: 7 options.
Total numbers = \( 9 \times 9 \times 8 \times 7 = 4536 \).
Q3
How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?
Answer:
Available digits: 1, 2, 3, 4, 6, 7 (Total 6).
Even digits: 2, 4, 6 (3 options).
- Unit's place: 3 ways (2, 4, or 6).
- Ten's place: 5 ways (remaining digits).
- Hundred's place: 4 ways.
Total numbers = \( 3 \times 5 \times 4 = 60 \).
Q4
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?
Answer:
(i) Total numbers:
\( ^5P_4 = 5 \times 4 \times 3 \times 2 = 120 \).
(ii) Even numbers:
Unit's digit must be 2 or 4 (2 options).
Remaining 3 places filled by remaining 4 digits: \( ^4P_3 = 24 \).
Total even numbers = \( 2 \times 24 = 48 \).
Q5
From a committee of 8 persons, in how many ways can we choose a chairman and a vice chairman assuming one person can not hold more than one position?
Answer:
We need to select 2 persons from 8 and arrange them (one Chairman, one Vice Chairman).
\( ^8P_2 = 8 \times 7 = 56 \) ways.
Q6
Find \( n \) if \( ^{n-1}P_3 : ^nP_4 = 1 : 9 \).
Answer:
\( \frac{^{n-1}P_3}{^nP_4} = \frac{1}{9} \)
\( \frac{\frac{(n-1)!}{(n-1-3)!}}{\frac{n!}{(n-4)!}} = \frac{1}{9} \implies \frac{(n-1)!}{(n-4)!} \times \frac{(n-4)!}{n!} = \frac{1}{9} \)
\( \frac{(n-1)!}{n(n-1)!} = \frac{1}{9} \implies \frac{1}{n} = \frac{1}{9} \)
\( n = 9 \).
Q7
Find \( r \) if (i) \( ^5P_r = 2 \cdot ^6P_{r-1} \) (ii) \( ^5P_r = ^6P_{r-1} \).
Answer:
(i) \( \frac{5!}{(5-r)!} = 2 \times \frac{6!}{(6-(r-1))!} \)
\( \frac{5!}{(5-r)!} = 2 \times \frac{6 \times 5!}{(7-r)!} \)
\( 1 = \frac{12}{(7-r)(6-r)} \)
\( r^2 - 13r + 42 = 12 \implies r^2 - 13r + 30 = 0 \)
\( (r-3)(r-10) = 0 \). Since \( r \le 5 \), \( r = 3 \).
(ii) \( \frac{5!}{(5-r)!} = \frac{6!}{(7-r)!} \)
\( 1 = \frac{6}{(7-r)(6-r)} \)
\( r^2 - 13r + 42 = 6 \implies r^2 - 13r + 36 = 0 \)
\( (r-4)(r-9) = 0 \). Since \( r \le 5 \), \( r = 4 \).
Q8
How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once?
Answer:
The word EQUATION has 8 distinct letters.
Number of permutations = \( ^8P_8 = 8! = 40320 \).
Q9
How many words 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?
(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?
Answer:
MONDAY has 6 distinct letters. Vowels: O, A (2). Consonants: M, N, D, Y (4).
(i) \( ^6P_4 = 6 \times 5 \times 4 \times 3 = 360 \).
(ii) \( ^6P_6 = 6! = 720 \).
(iii) First place: 2 options (O, A). Remaining 5 places: 5! ways.
Total = \( 2 \times 120 = 240 \).
Q10
In how many of the distinct permutations of the letters in MISSISSIPPI do the four I’s not come together?
Answer:
Total letters = 11 (M=1, I=4, S=4, P=2).
Total permutations = \( \frac{11!}{4!4!2!} = 34650 \).
Permutations where 4 I's come together: Treat (IIII) as 1 unit.
Units: {IIII}, M, S, S, S, S, P, P (Total 8 units).
Arrangements = \( \frac{8!}{4!2!} = 840 \).
Required = Total - Together = \( 34650 - 840 = 33810 \).
Q11
In how many ways can the letters of the word PERMUTATIONS be arranged if
(i) words start with P and end with S
(ii) vowels are all together
(iii) there are always 4 letters between P and S?
(i) words start with P and end with S
(ii) vowels are all together
(iii) there are always 4 letters between P and S?
Answer:
Word: PERMUTATIONS (12 letters). T repeats 2 times.
(i) P...S: Fix P and S. Remaining 10 letters (2 T's).
Ways = \( \frac{10!}{2!} = 1814400 \).
(ii) Vowels together: Vowels (E, U, A, I, O) = 5. Consonants = 7 (P, R, M, T, T, N, S).
Treat vowels as 1 block. Total entities = 7 + 1 = 8.
Arrangements = \( \frac{8!}{2!} \times 5! = 20160 \times 120 = 2419200 \).
(iii) 4 letters between P and S:
Positions for P and S: (1,6), (2,7), ..., (7,12). Total 7 positions. P and S can interchange (2 ways).
Remaining 10 letters arrange in \( \frac{10!}{2!} \) ways.
Total = \( 7 \times 2 \times \frac{10!}{2!} = 25401600 \).
Q1
If \( ^nC_8 = ^nC_2 \), find \( ^nC_2 \).
Answer:
Using property \( ^nC_x = ^nC_y \implies x + y = n \):
\( n = 8 + 2 = 10 \).
\( ^nC_2 = ^{10}C_2 = \frac{10 \times 9}{2} = 45 \).
Q2
Determine \( n \) if (i) \( ^{2n}C_3 : ^nC_3 = 12 : 1 \) (ii) \( ^{2n}C_3 : ^nC_3 = 11 : 1 \)
Answer:
(i) \( \frac{2n(2n-1)(2n-2)}{6} \times \frac{6}{n(n-1)(n-2)} = \frac{12}{1} \)
\( \frac{2(2n-1) \cdot 2(n-1)}{(n-1)(n-2)} = 12 \implies \frac{4(2n-1)}{n-2} = 12 \)
\( 2n-1 = 3(n-2) \implies 2n-1 = 3n-6 \implies n = 5 \).
(ii) \( \frac{4(2n-1)}{n-2} = 11 \implies 8n-4 = 11n-22 \)
\( 3n = 18 \implies n = 6 \).
Q3
How many chords can be drawn through 21 points on a circle?
Answer:
A chord connects 2 points.
\( ^{21}C_2 = \frac{21 \times 20}{2} = 210 \).
Q4
In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?
Answer:
Select 3 boys from 5: \( ^5C_3 = 10 \).
Select 3 girls from 4: \( ^4C_3 = 4 \).
Total ways = \( 10 \times 4 = 40 \).
Q5
Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour.
Answer:
3 Red from 6: \( ^6C_3 = 20 \).
3 White from 5: \( ^5C_3 = 10 \).
3 Blue from 5: \( ^5C_3 = 10 \).
Total ways = \( 20 \times 10 \times 10 = 2000 \).
Q6
Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination.
Answer:
Select 1 Ace from 4: \( ^4C_1 = 4 \).
Select 4 other cards from 48: \( ^{48}C_4 = 194580 \).
Total = \( 4 \times 194580 = 778320 \).
Q7
In how many ways can one select a cricket team of eleven from 17 players in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers?
Answer:
Bowlers: 5, Others: 12. Need 4 Bowlers, 7 Others.
\( ^5C_4 \times ^{12}C_7 = 5 \times 792 = 3960 \).
Q8
A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected.
Answer:
\( ^5C_2 \times ^6C_3 = 10 \times 20 = 200 \).
Q9
In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?
Answer:
2 courses are fixed. We need to choose \( 5-2=3 \) courses from the remaining \( 9-2=7 \).
\( ^7C_3 = 35 \).
Q1
How many words, with or without meaning, each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER?
Answer:
Vowels: A, U, E (3). Consonants: D, G, H, T, R (5).
Selection: \( ^3C_2 \times ^5C_3 = 3 \times 10 = 30 \).
Arrangement of 5 letters: \( 30 \times 5! = 30 \times 120 = 3600 \).
Q2
How many words can be formed using all the letters of the word EQUATION at a time so that the vowels and consonants occur together?
Answer:
Vowels (5): {A, E, I, O, U}. Consonants (3): {Q, T, N}.
Treat Vowels as 1 group, Consonants as 1 group. Arrangement of groups: \( 2! \).
Internal arrangements: \( 5! \) (vowels) and \( 3! \) (consonants).
Total = \( 2! \times 5! \times 3! = 2 \times 120 \times 6 = 1440 \).
Q3
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) atleast 3 girls ? (iii) atmost 3 girls ?
(i) exactly 3 girls ? (ii) atleast 3 girls ? (iii) atmost 3 girls ?
Answer:
(i) Exactly 3 girls: (3G, 4B) \( \to ^4C_3 \times ^9C_4 = 4 \times 126 = 504 \).
(ii) At least 3 girls: (3G, 4B) + (4G, 3B)
\( 504 + (^4C_4 \times ^9C_3) = 504 + (1 \times 84) = 588 \).
(iii) At most 3 girls: 0G, 1G, 2G, or 3G.
Total ways - (4 girls case) = \( ^{13}C_7 - 84 = 1716 - 84 = 1632 \).
Q4
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?
Answer:
Letters: A(2), E, I(2), M, N(2), O, T, X. Total 11.
Words starting with A: Fix A first. Remaining 10 letters have 2 I's, 2 N's.
Ways = \( \frac{10!}{2!2!} = 907200 \).
These are all words before E.
Q5
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?
Answer:
Divisible by 10 means unit digit is 0. Fix 0 at end.
Remaining 5 places filled by 1, 3, 5, 7, 9.
\( 5! = 120 \).
Q6
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?
Answer:
Selection: \( ^5C_2 \times ^{21}C_2 = 10 \times 210 = 2100 \).
Arrangement (4 letters): \( 2100 \times 4! = 50400 \).
Q7
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?
Answer:
Choices (Part I, Part II): (3,5), (4,4), (5,3).
- \( ^5C_3 \times ^7C_5 = 10 \times 21 = 210 \)
- \( ^5C_4 \times ^7C_4 = 5 \times 35 = 175 \)
- \( ^5C_5 \times ^7C_3 = 1 \times 35 = 35 \)
Total = \( 210 + 175 + 35 = 420 \).
Q8
Determine the number of 5-card combinations out of a deck of 52 cards if each selection of 5 cards has exactly one king.
Answer:
1 King from 4, 4 others from 48.
\( ^4C_1 \times ^{48}C_4 = 4 \times 194580 = 778320 \).
Q9
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?
Answer:
Positions: 1 2 3 4 5 6 7 8 9.
Women in even places (2, 4, 6, 8): \( 4! \) ways.
Men in odd places (1, 3, 5, 7, 9): \( 5! \) ways.
Total = \( 24 \times 120 = 2880 \).
Q10
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?
Answer:
Case 1: All 3 join. Choose remaining 7 from 22. \( ^{22}C_7 = 170544 \).
Case 2: None join. Choose 10 from 22. \( ^{22}C_{10} = 646646 \).
Total = \( 170544 + 646646 = 817190 \).
Q11
In how many ways can the letters of the word ASSASSINATION be arranged so that all the S’s are together?
Answer:
Letters: A(3), S(4), I(2), N(2), T(1), O(1). Total 13.
Treat (SSSS) as 1 unit. Total units = 13 - 4 + 1 = 10.
Arrangements = \( \frac{10!}{3!2!2!} = \frac{3628800}{6 \times 2 \times 2} = 151200 \).