# Inter 2nd Year Maths 2A Permutations and Combinations Solutions Ex 5(a)

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Permutations and Combinations Solutions Exercise 5(a) will help students to clear their doubts quickly.

## Intermediate 2nd Year Maths 2A Permutations and Combinations Solutions Exercise 5(a)

I.

Question 1.
If nP3 = 1320, find n.
Solution:
Hint: nPr = $$\frac{n !}{(n-r) !}$$ = n(n – 1)(n – 2)…….(n – r + 1)
nP3 = 1320
= 10 × 132
= 10 × 12 × 11
= 12 × 11 × 10
= 12P3
∴ n = 12

Question 2.
If nP7 = 42 . nP5, find n.
Solution:
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 – 5) (n – 6) = 42
⇒ (n – 5) (n – 6) = 7 × 6
⇒ n – 5 = 7 or n – 6 = 6
⇒ n = 12

Question 3.
If (n+1)P5 : nP6 = 2 : 7, find n.
Solution:
(n+1)P5 : nP6 = 2 : 7
⇒ $$\frac{(n+1)_{P_5}}{n_{P_6}}=\frac{2}{7}$$
⇒ $$\frac{(n+1) n(n-1)(n-2)(n-3)}{n(n-1)(n-2)(n-3)(n-4)(n-5)}=\frac{2}{7}$$
⇒ 7(n + 1) = 2(n – 4) (n – 5)
⇒ 7n + 7 = 2n2 – 18n + 40
⇒ 2n2 – 25n + 33 = 0
⇒ 2n2 – 22n – 3n + 33 = 0
⇒ 2n(n – 11) – 3(n – 11) = 0
⇒ (n – 11) (2n – 3) = 0
⇒ n = 11 or $$\frac{3}{2}$$
Since n is a positive integer
∴ n = 11

Question 4.
If 12P5 + 5 . 12P4 = 13Pr, find r.
Solution:
We have
(n-1)Pr + r . (n-1)P(r-1) = nPr and r ≤ n
12P5 + 5 . 12P4 = 13P5 = 13Pr (given)
⇒ r = 5

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

⇒ 18 × 7 = 9(19 – r)
⇒ 14 = 19 – r
⇒ r = 19 – 14 = 5

Question 6.
A man has 4 sons and there are 5 schools within his reach. In how many ways can he admit his sons into 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

II.

Question 1.
If there are 25 railway stations on a railway line, how many types of single second-class tickets must be printed, so as to enable a passenger to travel from one station to another?
Solution:
Number of stations on a railway line = 25
∴ The number of single second class tickets must be printed so as to enable a passenger to travel from one station to another = 25P2
= 25 × 24
= 600

Question 2.
In a class, there are 30 students. On New year’s day, every student posts a greeting card to all his/her classmates. Find the total number of greeting cards posted by them.
Solution:
The number of students in a class is 30.
∴ Total number of greeting cards posted by every student to all his/her classmates = 30P2
= 30 × 29
= 870

Question 3.
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:
Vowels – A, E, I, O, U
In a given, word, the number of vowels is 3
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

Question 4.
Find the sum of all 4 digited numbers that can be formed using the digits 0, 2, 4, 7, 8 without repetition.
Solution:
First Method:
The number of 4 digited numbers formed by using the digits 0, 2, 4, 7, 8 without repetition = 5P44P3
= 120 – 24
= 96
Out of these 96 numbers,
4P33P2 numbers contain 2 in the units place
4P33P2 numbers contain 2 in the tens place
4P33P2 numbers contain 2 in the hundreds place
4P3 numbers contain 2 in the thousands place

∴ The value obtained by adding 2 in all the numbers = (4P33P2) 2 + (4P33P2) 20 + (4P33P2) 200 + 4P3 × 2000
= 4P3 (2 + 20 + 200 + 2000) – 3P2 (2 + 20 + 200)
= 24 × (2222) – 6 × (222)
= 24 × 2 × 1111 – 6 × 2 × 111
Similarly, 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 × 111
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 × 111 × (2 + 4 + 7 + 8)
= 26664 (21) – 666 (21)
= 21 (26664 – 666)
= 21 × 25998
= 545958

Second Method:
If Zero is one among the given n digits, then the sum of the r – digited numbers that can be formed using the given ‘n’ distinct digits (r ≤ n ≤ 9) is
(n-1)P(r-1) × sum of the digits × 111 …… 1 (r times) – (n-2)P(r-2) × Sum of the digits × 111 ……… 1 [(r – 1) times]
Hence n = 5, r = 4, digits are {0, 2, 4, 7, 8}
Hence the sum of all 4 digited numbers that can be formed using the digits {0, 2, 4, 7, 8} without repetition is
(5-1)P(4-1) × (0 + 2 + 4 + 7 + 8) × (1111) – (5-2)P(4-2) × (0 + 2 + 4 + 7 + 8) × (111)
= 4P3 (21) × 1111 – 3P2 (21) (111)
= 24 × 21 × 1111 – 6 (21) (111)
= 21 (26664) – 21 (666)
= 21 (26664 – 666)
= 21 (25998)
= 545958

Question 5.
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 of 5 digits are greater than 4000.
In the 4-digit numbers, the number starting with 4 or 6 or 8 are greater than 4000.
The number of 4-digit numbers that 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

Question 6.
Find the number of ways of arranging the letters of the word MONDAY so that no vowel occupies an even place.
Solution:
In the word MONDAY, there are two vowels, 4 consonants and three even places, three odd places.
Since no vowel occupies an even place, the two vowels can be filled in the three odd places in 3P2 ways.
The 4 consonants can be filled in the remaining 4 places in 4! ways.
∴ The number of required arrangements = 3P2 × 4!
= 6 × 24
= 144

Question 7.
Find the number of ways of arranging 5 different mathematics books, 4 different Physics books, and 3 different chemistry books such that the books of the same subject are together.
Solution:
The number of ways of arranging Mathematics, Physics, and Chemistry books are arranged 3! ways.
5 different Mathematics books are arranged themselves in 5! ways.
4 different Physics books are arranged themselves in 4! ways.
3 different Chemistry books are arranged themselves in 3! ways.
∴ The number of required arrangements = 3! × 5! × 4! × 3!
= 6 × 120 × 24 × 6
= 1,03,680

III.

Question 1.
Find the number of 5-letter words that can be formed using the letters of the word CONSIDER. How many of them begin with “C”, how many of them end with ‘R’ and how many of them begin with “C” and end with “R”?
Solution:

The number of 5 letter words that can be formed using the letters of the word CONSIDER = 8P5
= 8 × 7 × 6 × 5 × 4
= 6720

The number of 5 letter words beginning with ‘C’ = 1 × 7P4
= 7 × 6 × 5 × 4
= 840

The number of 5 letter words end with ‘R’ = 1 × 7P4 = 840

The number of 5 letter words begins with ‘C’ and ends with ‘R’ = 1 × 1 × 6P3
= 6 × 5 × 4
= 120

Question 2.
Find the number of ways of seating 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:
A1, A2, A3, ……….., A10 are the ten students.
(i) Consider A1, A2, A3 as one unit and A4, A5, A6, A7, A8, A9, A10 as seven units.
These 8 units can be arranged in 8! ways.
The students A1, A2, A3 in one unit can be arranged among themselves in 3! ways.
∴ The number of ways seating 10 students in which A1, A2, A3 sit together = (8!) (3!)

(ii) To arrange A1, A2, A3 to sit in a specified order,
A1, A2, A3 can arrange in 10 positions in specific order in $$\frac{{ }^{10} P_3}{3 !}$$ ways.
The remaining 7 people 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 that order as one unit.
Now there are 8 objects, they can be arranged in 8! ways.
∴ The number of ways of A1, A2, A3 sit together in specified order = 8! ways

Question 3.
Find the number of ways in which 5 red balls, 4 black balls of different sizes can be arranged in a row so that
(i) no two balls of the same colour come together.
(ii) the balls of the same colour come together.
Solution:
No. of red balls = 5
No. of black balls = 4
(i) No. two balls of the same colour come together.
For the required arrangements first, we arrange 4 black balls it can be done in 4! ways.

There are 5 places to arrange 5 red balls it can be done in 5! ways.
∴ The number of required arrangements = 4! 5!

(ii) The balls of the same colour come together.
For the required arrangements 4 black balls of different sizes can be considered as one object and 5 red balls can be considered as one object. These can be arranged in 2! ways.
The 4 black balls can be permuted among themselves in 4! ways.
The 5 red balls can be permuted among themselves in 5! ways.
∴ The no. of required arrangements = 2! 4! 5!

Question 4.
Find the number of 4-digit numbers that can be formed using the digits 1, 2, 5, 6, 7. How many of them are divisible by
(i) 2
(ii) 3
(iii) 4
(iv) 5
(v) 25
Solution:
The number of 4 digited numbers that can be formed using the digits 1, 2, 5, 6, 7 is 5P4 = 120.
(i) A number is divisible by 2 when its unit place must be filled with an even digit from among the given integers. This can be done in 2 ways.

Now, the remaining 3 places can be filled with the remaining 4 digits in 4P3 = 4 × 3 × 2 = 24 ways.
∴ The number of 4 digited numbers divisible by 2 = 2 × 24 = 48

(ii) A number is divisible by 3 only when the sum of the digits in that number is a multiple of 3.
Sum of the given 5 digits = 1 + 2 + 5 + 6 + 7 = 21
The 4 digits such that their sum is a multiple of 3 from the given digits are 1, 2, 5, 7 (sum is 15)
They can be arranged in 4! ways and all these 4 digited numbers are divisible by 3.
∴ The number of 4 digited numbers divisible by 3 = 4! = 24

(iii) A number is divisible by 4 only when the last two places (tens and units places) of it are a multiple of 4.

Hence the last two places should be filled with one of the following 12, 16, 52, 56, 72, 76.
Thus the last two places can be filled in 6 ways.
The remaining two places can be filled by the remaining 3 digits in 3P2 = 3 × 2 = 6 ways.
∴ The number of 4 digited numbers divisible by 4 = 6 × 6 = 36.

(iv) A number is divisible by 5 when its units place must be filled by 5 from the given integers 1, 2, 5, 6, 7. This can be done in one way.

The remaining 3 places can be filled with the remaining 4 digits in 4P3 = 4 × 3 × 2 = 24 ways.
∴ The number of 4 digited numbers divisible by 5 = 1 × 24 = 24

(v) A number is divisible by 25 when its last two places are filled with either 25 or 75.

Thus the last two places can be filled in 2 ways.
The remaining 2 places from the remaining 3 digits can be filled in 3P2 = 6 ways.
∴ The number of 4 digited numbers divisible by 25 = 2 × 6 = 12

Question 5.
If the letters of the word MASTER are permuted in all possible ways and the words thus formed are arranged in the dictionary order, then find the ranks of the words
(i) REMAST
(ii) MASTER
Solution:
(i) The alphabetical order of the letters of the given word is A, E, M, R, S, T
The number of words beginning with A is 5! = 120
The number of words that begins with E is 5! = 120
The number of words begins with M is 5! = 120
The number of words beginning with RA is 4! = 24
The number of words beginning with REA is 3! = 6
The next word is REMAST
∴ Rank of the word REMAST = 3(120) + 24 + 6 + 1
= 360 + 31
= 391

(ii) The alphabetical order of the letters of the given word is A, E, M, R, S, T
The number of words beginning with A is 5! = 120
The number of words that begins with E is 5! = 120
The number of words beginning with MAE is 3! = 6
The number of words that begins with MAR is 3! = 6
The number of words beginning with MASE is 2! = 2
The number of words begins with MASR is 2! = 2
The next word is MASTER.
∴ Rank of the word MASTER = 2(120) + 2(6) + 2(2) + 1
= 240 + 12 + 4 + 1
= 257

Question 6.
If the letters of the word BRING are permuted in all possible ways and the words thus formed are arranged in the dictionary order, then find the 59th word.
Solution:
By using the letters of the word BRING in alphabetical order the 59th word must start with ‘I’.
The next word is 59th = IGRBN

Question 7.
Find the sum of all 4 digited numbers that can be formed using the digits 1, 2, 4, 5, 6 without repetition.
Solution:
The number of 4 digited numbers formed by using the digits 1, 2, 4, 5, 6 without repetition = 5P4 = 120
Out of these 120 numbers,
4P3 numbers contain 2 in the units place
4P3 numbers contain 2 in the tens place
4P3 numbers contain 2 in the hundreds place
4P3 numbers contain 2 in the thousands place

∴ The value obtained by adding 2 in all the numbers = 4P3 × 2 + 4P3 × 20 + 4P3 × 200 + 4P3 × 2000
= 4P3 (2 + 20 + 200 + 2000)
= 4P3 (2222)
= 4P3 × 2 × 1111
Similarly, the value obtained by adding 1 is 4P3 × 1 × 1111
the value obtained by adding 4 is 4P3 × 4 × 1111
the value obtained by adding 5 is 4P3 × 5 × 1111
the value obtained by adding 6 is 4P3 × 6 × 1111
∴ The sum of all the numbers = 4P3 × 1 × 1111 + 4P3 × 2 × 1111 + 4P3 × 4 × 1111 + 4P3 × 5 × 1111 + 4P3 × 6 × 1111
= 4P3 (1111) (1 + 2 + 4 + 5 + 6)
= 24 (1111) (18)
= 4,79,952

Second Method:
The sum of the r-digited numbers that can be formed using the given ‘n’ distinct non-zero digits (r ≤ n ≤ 9) is (n-1)P(r-1) × sum of all digits × 111 …… 1 (r times)
Hence n = 5, r = 4, digits = {1, 2, 4, 5, 6}
The sum of all 4 digited numbers that can be formed using the digits 1, 2, 4, 5, 6 without repetition is (5-1)P(4-1) × (1 + 2 + 4 + 5 + 6) × (1111)
= 4P3 (18) (1111)
= 24 × 18 × 1111
= 4,79,952

Question 8.
There are 9 objects and 9 boxes. Out of 9 objects, 5 cannot fit in three small boxes. How many arrangements can be made such that each object can be put in one box only?
Solution:
No. of objects = 9
No. of boxes = 9
For the required arrangements, out of 9 objects, 5 cannot bit in three small boxes.
These five can be arranged in 6 boxes these can be done in 6P5 ways.
The remaining 4 objects can be arranged in the remaining 4 boxes it can be done in 4! ways.
∴ No. of required arrangements = 6P5 × 4! ways