Inter 1st Year Maths Permutations and Combinations Solutions Exercise 6c

Practicing the AP Board Solutions Class 11 Maths and Chapter 6 Inter 1st Year Maths Permutations and Combinations Solutions Exercise 6c Pdf Download will help students to clear their doubts quickly.

Intermediate 1st Year Maths Permutations and Combinations Solutions Exercise 6c

Permutations and Combinations Exercise 6c Solutions

Permutations and Combinations Class 11 Exercise 6c Solutions – Permutations and Combinations 6c Exercise Solutions

I.

Question 1.
If ⁿP₃ = 1320, find n.
Solution:
ⁿP₃ = 1320
ⁿP₃ = 10 × 132
= 10 × 12 × 11
= 12 × 11 × 10
= ¹²P₃
∴ n = 12

Question 2.
If ⁿP₇ = 42 . ⁿP₅, find n.
Solution:
Given ⁿP₇ = 42 ⁿP₅
⇒ n(n – 1) (n – 2) (n – 3) (n – 4) (n – 5) (n – 6) = 42 n(n – 1) (n – 2) (n – 3) (n – 4)
⇒ n(n – 5) (n – 6) = 42
⇒ n² – 11n + 30 = 42
⇒ n² – 11n + 30 – 42 = 0
⇒ n² – 11n – 12 = 0
⇒ n² – 12n + n – 12 = 0
⇒ n(n – 12) + 1(n – 12) = 0
⇒ (n – 12) (n + 1) = 0
⇒ n = -1, n = 12
As n cannot be negative, n = 12.

Question 3.
⁽ⁿ⁺¹⁾P₅ : ⁿP₆ = 2 : 7, find n.
Solution:
⁽ⁿ⁺¹⁾P₅ : ⁿP₆ = 2 : 7
⇒ 7 ⁽ⁿ⁺¹⁾P₃ = 2 ⁿP₆
⇒ 2n(n – 1) (n – 2) (n – 3) (n – 4) (n – 5) = 7(n + 1) n (n – 1) (n – 2) (n – 3)
⇒ 2(n – 4) (n – 5) = 7(n + 1)
⇒ 2(n² – 9n + 20) = 7n + 7
⇒ 2n² – 18n + 40 = 7n + 7
⇒ 2n² – 25n + 33 = 0
⇒ 2n² – 22n – 3n + 33 = 0
⇒ 2n(n – 11) – 3(n – 11) = 0
⇒ (n – 11)(2n – 3) = 0
⇒ n = 11 (∴ n cannot equal 3/2).

Question 4.
If ¹²P₅ + 5 ¹²P₄ = ¹³P_r, find r.
Solution:

Question 5.
If ¹⁸P_(r-1) : ¹⁷P_(r-1) = 9 : 7, find r.
Solution:

Question 6.
Find n if ^n-1P₃ : ⁿP₄ = 1 : 9.
Solution:
The given equation can be written as
\(\frac{{ }^{n-1} P_3}{{ }^n P_4}=\frac{1}{9}\)
By substituting the values, we get
\(\frac{\frac{(n-1)!}{(n-4)!}}{\frac{n!}{(n-4)!}}=\frac{1}{9}\)
The simplification
⇒ \(\frac{(n-1)!}{n!}=\frac{1}{9}\)
⇒ \(\frac{(n-1)!}{n \times(n-1)!}=\frac{1}{9}\)
⇒ \(\frac{1}{n}=\frac{1}{9}\)
⇒ n = 9

Question 7.
Find r if
(i) ⁵P_r = 2 ⁶P_r-1
(ii) ⁵P_r = ⁶P_r-1
Solution:
(i) Substituting the values, we get
\(\frac{5!}{(5-r)!}=2 \frac{6!}{(7-r)!}\)
The above equation can be written as
\(\frac{(7-r)!}{(5-r)!}=2 \frac{6!}{5!}\)
The simplifying we got
⇒ (7 – r)(6 – r) = 2(6)
⇒ 42 – 13r + r² = 12
⇒ r² – 13r + 30 = 0
⇒ r² – 10r – 3r + 30 = 0
⇒ r(r – 10) – 3(r – 10) = 0
⇒ (r – 3) (r – 10) = 0
⇒ r = 3, or r = 10
But r = 10 is rejected as in ⁵Pr, r cannot be greater than 5.
Therefore, r = 3.

(ii) The above equation can be written as
\(\frac{5!}{(5-r)!}=\frac{6!}{(7-r)!}\)
⇒ \(\frac{(7-r)!}{(5-r)!}=\frac{6!}{5!}\)
⇒ (7 – r)(6 – r) = 6
⇒ 42 – 13r + r² = 6
⇒ r² – 13r + 36 = 0
⇒ r² – 9r – 4r + 36 = 0
⇒ r(r – 9) – 4(r – 9) = 0
⇒ (r – 4) (r – 9) = 0
⇒ r = 4 or r = 9
But r = 9 is rejected as in ⁵Pr, r cannot be greater than 5.
Therefore, r = 4.

Question 8.
A man has 4 sons, and there are 5 schools within his reach. In how many ways can he admit his sons to the schools so that no two of them will be in the same school?
Solution:
The number of ways of admitting 4 sons into 5 schools if no two of them will be in the same school = ⁵P₄
= 5 × 4 × 3 × 2
= 120

Question 9.
How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?
Solution:
Total number of any digits possible for choosing = 9
Number of places for which a digit has to be taken = 3
As there is no repetition allowed,
No. of permutations = ⁹P₃
= \(\frac{9!}{(9-3)!}\)
= \(\frac{9 \times 8 \times 7 \times 6!}{6!}\)
= 504

Question 10.
How many 4-digit numbers are there with no digit repeated?
Solution:
To find the 4-digit number (digits do not repeat), we will have 4 places where 4 digits are to be put.
So, at the thousand’s place = There are 9 ways, as cannot be at the thousand’s place = 9 ways.
At the hundredth’s place = There are 9 digits to be filled, as 1 digit is already taken = 9 ways.
At the unit’s place, 7 digits can be filled in 7 ways.
The total number of ways to fill the four places = 9 × 9 × 8 × 7 = 4536
So, a total of 4536 four-digit numbers can be there with no digits repeated.

Question 11.
How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?
Solution:
An even number means that the last digit should be even
The number of possible digits at One’s place = 3 (2, 4, and 6)
Number of permutations = ³P₁
= \(\frac{3!}{(3-1)!}\)
= 3
One of the digits is taken at one’s place; the number of possible digits available = 5
Number of permutations = ⁵P₂
= \(\frac{5!}{(5-2)!}\)
= \(\frac{5 \times 4 \times 3!}{3!}\)
= 20
Therefore, the total number of Permutations = 3 × 20 = 60

Question 12.
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?
Solution:
Total number of digits possible for choosing = 5
Number of places for which a digit has to be taken = k
As there is no repetition allowed.
⁵P₄ = \(\frac{5!}{(5-4)!}\) = 120
The number will be even when 2 and 4 are in one’s place.
The possibility of (2, 4) at one’s place = \(\frac {2}{5}\) = 0.4
The total number of even number = 120 × 0.4 = 48

Question 13.
From a committee of 8 persons, in how many ways can we choose a chairman and a vice chairman, assuming that one person cannot hold more than one position?
Solution:
Total number of people in committee = 8
Number of positions to be filled = 2
Number of positions is = ⁸P₂
= \(\frac{8!}{(8-2)!}\)
= \(\frac{8!}{6!}\)
= 56

Question 14.
Find the number of 4-digit numbers that can be formed using the digits 1, 2, 4, 5, 7, 8 when repetition is allowed.
Solution:
Given the digits 1, 2, 4, 5, 7, 8
There are 6 digits in total, and we need to form 4-digit numbers using repetition allowed.
For each of the 4 places (thousands, hundreds, tens, units), we use any of the 6-digit
Total number = 6 × 6 × 6 × 6 = 6⁴ = 1296

Question 15.
Find the number of 5-letter words that can be formed using the letters of the word RHYME if each letter can be used any number of times.
Solution:
The number of 5-letter words that can be formed using the letters of the word RHYME if each letter can be used any number of times = 5⁵ = 3125

Question 16.
Find the number of functions from a set A containing 5 elements into a set B containing 4 elements.
Solution:
The number of functions = n(B)^n(A)
= 4⁵
= 1024

Question 17.
Find the number of ways of arranging the letters of the words.
(i) MATHEMATICS
(ii) SINGING
(iii) PERMUTATION
(iv) COMBINATION
(v) INTERMEDIATE
Solution:
(i) The word MATHEMATICS contains 11 letters in which there are 2M’s that are alike, 2A’s that are alike, 2T’s that are alike, and the rest are different.

(ii) The word SINGING contains 7 letters in which there are 2I’s that are alike, 2N’s that are alike, 2G’s that are alike, and the rest are different.
∴ The number of required arrangements = \(\frac{7!}{2!2!2!}\)

(iii) The word PERMUTATION contains 11 letters in which there are 2T’s that are alike, and the rest are different.
∴ The Number of required arrangements = \(\frac{(11)!}{2!}\)

(iv) The word COMBINATION contains 11 letters in which there are 2O’s that are alike, 2I’s that are alike, 2 N’s that are alike, and the rest are different.
∴ The number of required arrangements = \(\frac{(11)!}{2!2!2!}\)

(v) The word INTERMEDIATE contains 12 letters in which there are 2I’s that are alike, 2T’s that are alike, 3E’s that are alike, and the rest are different.
∴ The number of required arrangements = \(\frac{(12)!}{2!2!3!}\)

Question 18.
In how many of the distinct permutations of the letters in MISSISSIPPI do the four I’s not come together?
Solution:
Total number of letters in MISSISSIPPI = 11 letters,
number of occurrences

Number of permutations = \(\frac{11!}{1!4!2!}\) = 34650
We have taken that 4I’s come together and they are treated as 1 letter
∴ Total number of letters = 11 – 4 + 1 = 8
Number of permutations = \(\frac{8!}{1!4!2!}\) = 840
Therefore total number of permutations where four I’s don’t come together = 34650 – 840 = 33810

Question 19.
If there are 25 railway stations on a railway line, how many types of single second-class tickets must be printed to enable a passenger to travel from one station to another?
Solution:
No. of stations on a railway line = 25
∴ Number of single’ second class ticket must be printed to enable a passenger to travel from one station to another = ⁵P₂
= 25 × 24
= 600

Question 20.
Find the number of ways of arranging the letters of the word TRIANGLE so that the relative positions of the vowels and consonants are not disturbed.
Solution:
In the given word TRLANGLE, the number of vowels is 3, and the number of consonants is 5.
Since the relative positions of the vowels and consonants are not disturbed, the 3 vowels can be arranged in their relative positions in 3! ways, and the 5 consonants can be arranged in their relative positions in 5! ways.
∴ The number of required arrangements = (3!) (5!)
= (6) (120)
= 720

II.

Question 1.
How many words, with or without meaning, can be made from the letters of the word MONDAY, assuming that no letter is repeated?
(i) 4 letters are used at a time.
(ii) all letters are used at a time.
(iii) all letters are used, but the first letter is a vowel?
Solution:
(i) Number of letters to be used = 4
Number of permutations = ⁶P₄
= \(\frac{6!}{(6-4)!}\)
= \(\frac{6!}{2!}\)
= 360

(ii) Number of letters to be used = 6
Number of permutations = ⁶P₆
= \(\frac{6!}{(6-6)!}\)
= \(\frac{6!}{0!}\)
= 720

(iii) Number of vowels in MONDAY = 2 (O and A)
Number of permutations in vowel = ²P₁ = 2
Now, the remaining place = 5
Remaining letters to be used = ⁵P₅
= \(\frac{5!}{(5-5)!}\)
= \(\frac{5!}{0!}\)
= 120
The total number of permutations = 2 × 120 = 240

Question 2.
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) Are there always 4 letters between P and S?
Solution:
Total letters in PERMUTATIONS = 12
Repeating letter T = 2
(i) 1st position = P, 12th position = S
So we are left with 10 letters to arrange in the middle E, R, M, U, T, A, T, I, O, N (10 letters with T repeating twice) = \(\frac{10!}{2!}\)
= \(\frac {3628800}{2}\)
= 1814400

(ii) Vowels = E, U, A, I, O = 5 Vowels
All distinct Consonants: P, R, M, T, T, N, S = 7
Now we arrange 1 vowel block, 7 consonants (including 2T’s) = \(\frac{8!}{2!}\)
= \(\frac {40320}{2}\)
= 20160
All vowels are distinct = 5! =120 ways.
Total = 20160 × 120 = 2419200

(iii) We are to count the number of permutations such that there are exactly 4 letters between P and 8.
2 ways for each of the 7 position pairs
Total P-S position choices = 7 × 2 = 14
Ways to arrange remaining 10 letters = \(\frac{10!}{2!}\) = 4814400
Total = 14 × 1814400 = 25401600 ways.

Question 3.
Find the sum of all 4-digit numbers that can be formed using the digits 0, 2, 4, 7, 8 without repetition.
Solution:
The number of 4-digit numbers formed by using the digits 0, 2, 4, 7, 8 without repetition = ⁵P₄ – ⁴P₃
= 120 – 24
= 96
Out of these 96 numbers, ⁴P₃ – ³P₂ numbers contain 2 in the units place, ⁴P₃ – ³P₂ numbers contain 2 in the tens place, ⁴P₃ – ³P₂ numbers contain 2 in the hundreds place, and ⁴P₃ numbers contain 2 in the thousands place.
∴ The value obtained by adding 2 in all the numbers = (⁴P₃ – ³P₂) 2 + (⁴P₃ – ³P₂) 20 + (⁴P₃ – ³P₂) 200 + ⁴P₃ × 2000
= ⁴P₃ (2 + 20 + 200 + 2000) – 3P2 (2 + 20 + 200)
= 24 × (2222) – 6(222)
= 24 × 2 × 1111 – 6 × 2 × 111
The value obtained by adding 4 is 24 × 4 × 1111 – 6 × 4 × 111
The value obtained by adding 7 is 24 × 7 × 1111 – 6 × 7 × 1111
The value obtained by adding 8 is 24 × 8 × 1111 – 6 × 8 × 111
∴ The sum of all the numbers = (24 × 2 × 1111 – 6 × 2 × 111) + (24 × 4 × 1111 – 6 × 4 × 111) + (24 × 7 × 1111 – 6 × 7 × 111) + (24 × 8 × 1111 – 6 × 8 × 111)
= 24 × 1111 × (2 + 4 + 7 + 8) – 6 × 1111 × (2 + 4 + 7 + 8)
= 26664 (21) – 666 (21)
= 21(26664 – 666)
= 21(25998)
= 545958

Question 4.
Find the number of numbers that are greater than 4000 which can be formed using the digits 0, 2, 4, 6, 8 without repetition.
Solution:
While forming any digit number with the given digits, zero cannot be filled in the first place.
We can fill the first place with the remaining 4 digits.
The remaining places can be filled with the remaining 4 digits.
All the numbers with 5 digits are greater than 4000.
In the 4-digit numbers, the numbers starting with 4 or 8 are greater than 4000.
The number of 4-digit numbers which begin with 4 or 6, or 8 = 3 × ⁴P₃
= 3 × 24
= 72
The number of 5-digit numbers = 4 × 4!
= 4 × 24
= 96
∴ The number of numbers greater than 4000 is 72 + 96 = 168.

III.

Question 1.
Find the number of ways of arranging 10 students A₁, A₂, ………., A₁₀ in a row such that
(i) A₁, A₂, A₃ sit together.
(ii) A₁, A₂, A₃ sit in a specified order.
(iii) A₁, A₂, A₃ sit together in a specified order.
Solution:
(i) A₁, A₂, A₃, …….., A₁₀ are the ten students.
Consider A₁, A₂, A₃ as one person and A₄, A₅, A₆, A₇, A₈, A₉, A₁₀ as seven persons.
These 8 persons can be arranged in 8! ways.
The students A₁, A₂, and A₃ can be arranged among themselves in 3! ways.
∴ The number of ways of seating 10 students in which A₁, A₂, and A₃ sit together = (8!) (3!)

(ii) To arrange A₁, A₂, A₃ in a specified order, A₁, A₂, A₃ can be arranged in 10 positions in a specific order in \(\frac{{ }^{10} \mathrm{P}_3}{3!}\) ways.
The remaining 7 persons can be arranged in the remaining places in 7! ways.
∴ The number of ways of A₁, A₂, A₃ sit in a specific order = \(\frac{10!}{7!\times 3!} \times 7!\)
= \(\frac{10!}{3!}\)
= ¹⁰P₇

(iii) To arrange A₁, A₂, A₃ sit together in a specified order.
Consider A₁, A₂, A₃ in uicii order as one unit.
Now there are 8 objects; they can be arranged in 8! ways.
∴ The number of ways A₁, A₂, and A₃ sit together in a specified order = 8! ways.

Question 2.
Find the sum of all 4 digital numbers that can be formed using the digits 1, 2, 4, 5, 6 without repetition.
Solution:
5-digits total: 1, 2, 4, 5, 6
Forming 4-digit numbers without repetition, we need to select and arrange 4 digits from 5-digit numbers.
Number of such numbers = ⁵P₄
= 5 × 4 × 3 × 2
= 120
There are 4 positions: thousands, hundreds, tens, units.
Each of the 5-digits appears in each position exactly \(\frac {120}{4}\) = 30 items.
Let’s small digits = 1 + 2 + 4 + 5 + 6 = 18
Now, for each position, the sum of digits in that position = 30 × (1 + 2 + 4 + 5 + 6)
= 30 × 18
= 540
So, total sum = 540 × [1000 + 100 + 10 + 1]
= 540 × 1111
= 599940

Question 3.
9 different letters of the alphabet are given. Find the number of 4-letter words that can be formed using these 9 letters, which have
(i) No letter is repeated.
(ii) atleast one letter is repeated.
Solution:
The number of 4-letter words can be formed using the 9 different letters of the alphabet when repetition is allowed = 9⁴.
(i) The number of 4-letter words can be formed using the 9 different letters of an alphabet, in which no letter is repeated = ⁹P₄
= 9 × 8 × 7 × 6
= 3024

(ii) The number of 4-letter words can be formed using the 9 different letters of an alphabet, in which atleast one letter is repeated = 9⁴ – ⁹P₄
= 6561 – 3024
= 3537

Question 4.
Find the number of 4-digit numbers that can be formed using the digits 0, 1, 2, 3, 4, 5, which are divisible by 6 when repetition of the digits is allowed.
Solution:
The first place of the number can be filled by any one of the given digits except ‘0’ in 5 ways.
The 2nd and 3rd places can be filled by any one of the given 6 digits in 62 ways.

After filling up the first 3 places, if we fill the units place with the given 6 digits, we get 6 consecutive positive integers.
Out of these 6 consecutive integers, exactly one will be divisible by 6.
Hence, the unit’s place can be filled in one way.
∴ The number of 4-digit numbers formed using the given digits, which are divisible by 6 when repetition is allowed, is 5 × 6² × 1 = 180.

Question 5.
Find the number of ways of arranging the letters of the word ASSOCIATIONS. In how many of them
(i) all three S’s come together.
(ii) the two A’s do not come together.
Solution:
The given word ASSOCIATIONS has 12 letters in which there are 2A’s that are alike, 3S’s that are alike, 2O’s that are alike, 2I’s that are alike, and the rest are different.
∴ They can be arranged in \(\frac{(12)!}{2!3!2!2!}\) ways.
(i) Treat the 3 S’as one unit.
Then we have 9 + 1 = 10 entities in which there are 2 A’s that are alike, 2 0’s that are alike, 2 I’s that are alike, and the rest are different.
They can be arranged in \(\frac{(10)!}{2!2!2!}\) ways.
The 3 S’s among themselves can be arranged in \(\frac{3!}{3!}\) = 1 way.
∴ The number of required arrangements = \(\frac{(10)!}{3!2!2!}\)

(ii) Since 2 A’s do not come together, first arrange the remaining 10 letters in which there are 3 S’s are alike, 2 O’s are alike, 2 I’s are alike, and rest are different in \(\frac{(10)!}{3!2!2!}\) ways.
Then we can find 11 gaps between them.
The 2 A’s can be arranged in these 11 gaps in \(\frac{{ }^{11} \mathrm{P}_2}{2!}\) ways.
∴ The number of required arrangements = \(\frac{(10)!}{3!2!2!} \times \frac{{ }^{11} \mathrm{P}_2}{2!}\)