Balbharati Maharashtra State Board11th Commerce Maths Solution Book PdfChapter 6 Permutations and Combinations Ex 6.4 Questions and Answers.
Maharashtra State Board 11th Commerce Maths Solutions Chapter 6 Permutations and Combinations Ex 6.4
Solution & Step-by-Step Answer:
(i) There are 5 letters in the word DIVYA which can be arranged in 5! Way = 120 ways
(ii) There are 9 letters in the word SHANTARAM in which ‘A’ repeats 3 times.
∴ Number of permutations of the letters of the word SHANTARAM =
= 9 × 8 × 7 × 6 × 5 × 4
= 60480
(iii) There are 9 letters in the word REPRESENT in which ‘E’ repeats 3 times and ‘R’ repeats 2 times.
∴ Number of permutations of the letters of the word REPRESENT =
=
= 30240
(iv) There are 7 distinct letters in the word COMBINE which can be arranged among themselves in 7! = 5040 ways
Solution & Step-by-Step Answer:
There are total 9 books to be arranged on a shelf. Out of these 9 books, 2 books on English, 3 books on Hindi and 4 books on Mathematics are identical. ∴ Total number of arrangements = = = 9 × 4 × 7 × 5 = 1260 ∴ In 1260 distinct ways the books can be arranged on a shelf.
Solution & Step-by-Step Answer:
A coin is tossed 8 times. All heads are identical and all tails are identical. (i) We can obtain 4 heads and 4 tails in = = 70 ways ∴ In 70 different ways we can obtain 4 heads and 4 tails.
(ii) When at least 6 heads are to be obtained
∴ Outcome can be (6 heads and 2 tails) or (7 heads and 1 tail) or (8 heads)
∴ Number of ways in which it can be obtained =
= + 8 + 1
= 28 + 8 + 1
= 37
∴ In 37 different ways we can obtain at least 6 heads.
Solution & Step-by-Step Answer:
There is a total of 13 marbles in a bag. Out of these 5 are Red, 4 Blue, and 4 are Green marbles. All balls of the same colour are taken to be identical. ∴ Required number of arrangements =
Solution & Step-by-Step Answer:
There are 12 letters in the word MATHEMATICAL in which ‘M’ repeats 2 times, ‘A’ repeats 3 times, and ‘T’ repeats 2 times. ∴ Total number of arrangements = When all the vowels i.e., ‘A’, ‘A’, ‘A’, ‘E’, T are to be kept together Number of arrangements of these vowels = ways. Let us consider these vowels together as one unit. This unit is to be arranged with 7 other letters in which ‘M’ and ‘T’ repeated 2 times each. ∴ Number of arrangements = ∴ Total number of arrangements =
Solution & Step-by-Step Answer:
There are 11 letters in the word MAHARASHTRA in which ‘A’ is repeated 4 times, ‘H’ repeated 2 times, and ‘R’ repeated 2 times. ∴ Total number of arrangements is ∴ different words can be formed from the letters of the word MAHARASHTRA. (i) Other than M and T. there are 9 letters in which A repeats 4 times, H repeats twice, R repeats twice The number of arrangements of the a letter = These 9 letters create 10 gaps in which M and T are to be arranged The number of arrangements of M and T = 10P2 ∴ Total number arrangement having M and T never together =
(ii) When all vowels are together.
There are 4 vowels in the word MAHARASHTRA i.e., A, A, A, A
Let us consider these 4 vowels as one unit, they themselves can be arranged in = 1 way.
This unit is to be arranged with 7 other letters which can be done in 8! ways
∴ Total number of arrangements =
∴ different words can be formed if vowels are always together.
Solution & Step-by-Step Answer:
When ‘R’ is used thrice, ‘S’ is used twice and ‘T’ is used twice, ∴ Total number of letters available = 7, of which ‘S’ and ‘T’ repeat 2 times each, ‘R’ repeats 3 times. ∴ Required number of arrangements = = = 7 × 6 × 5 = 210 ∴ 210 different words can be formed with the letter R is used thrice and letters S and T are used twice each.
Solution & Step-by-Step Answer:
There are 6 letters in the word MUMBAI. These letters are to be arranged in such a way that ‘B’ is always next to ‘A’. Let us consider AB as one unit. This unit with other 4 letters in which ‘M’ repeats twice, is to be arranged. ∴ Total number of arrangements when B is always next to A = = = 60
Solution & Step-by-Step Answer:
There are 12 letters in the word CONSTITUTION, in which ‘O’, ‘N’, T repeat two times each, ‘T’ repeats 3 times. The arrangement starts and ends with ‘N’, 10 letters other than N can be arranged between two N, in which ‘O’ and ‘I’ repeat twice each and ‘T’ repeats 3 times. ∴ Total number of arrangements with the letter N at the beginning and at the end =
Solution & Step-by-Step Answer:
(i) There are 7 letters in the word ARRANGE in which A is repeated 2 times and R is repeated 2 times ∴ The number of arrangements = = 1260
(ii) A: set of words having 2A together
B: set of words having 2R together
Number of words having both A and both R not together
= 1260 – n(A ∪ B)
= 1260 – [n(a) + n(B) – n(A ∩ B)] ……(i)
n(A) = number of ways in which (AA) R, R, N, G, E are to be arranged
∴ n(A) = = 360
n(B) = number of ways in which (RR), A, A, N, G, E are to be arranged
∴ n(B) = = 360
n(A ∩ B) = number of ways in which (AA), (RR), N, G, E are to be arranged
∴ n(A ∩ B) = 5! = 120
Substituting n(A), n(B), n(A ∩ B) in (i), we get
Number of words having both A and both R not together
= 1260 – [360 + 360 – 120]
= 1260 – 600
= 660
Solution & Step-by-Step Answer:
5 digit numbers are to be formed from 2, 3, 2, 3, 4, 5. Case I: Numbers formed from 2, 2, 3, 4, 5 OR 2, 3, 3, 4, 5 Number of such numbers = × 2 = 5! = 120
Case II: Numbers formed from 2, 2, 3, 3 and any one of 4 or 5
Number of such numbers = × 2 = 60
Required number = 120 + 60 = 180
∴ 180 distinct 5 digit numbers can be formed using the digit 3, 2, 3, 2, 4, 5.
Solution & Step-by-Step Answer:
A number is to be formed with digits 3, 4, 5, 6, 7, 8, 9 such that odd digits always occupy the odd places. There are 4 odd digits i.e. 3, 5, 7, 9. They can be arranged at 4 odd places among themselves in 4! ways = 24 ways 3 even places of the number are occupied by even digits (i.e. 4, 6, 8). ∴ They can be arranged in 3! ways = 6 ways ∴ Total number of arrangements = 24 × 6 = 144 ∴ 144 numbers can be formed so that odd digits always occupy the odd positions.
Solution & Step-by-Step Answer:
A 6-digit number is to be formed using digits of 659942, in which 9 repeats twice. ∴ Total number of arrangements = = = 360 ∴ 360 different 6-digit numbers can be formed. For a number to be divisible by 2, Last digits can be selected in 3 ways Remaining 5 digit in which, 9 appears twice are arranged in ways ∴ Total number of arrangements = × 3 = 180 ∴ 180 numbers are divisible by 2.
Solution & Step-by-Step Answer:
There are 6 letters in the word INDIAN in which I and N repeat twice. Number of different words that can be formed using the letters of the word INDIAN = = = 180 ∴ 180 different words can be formed with the letters of the word INDIAN. When two N’s are together. Let us consider the two N’s as one unit. They can be arranged with 4 other letters in = = 60 ways. ∴ 2N can be arranged in 1 way ∴ Total number of arrangements = 60 × 1 = 60 ways ∴ 60 words are such that two N’s are together.
Solution & Step-by-Step Answer:
(i) When the two O’s are never together: Let us arrange the other 5 letters first, which can be done in 5! = 120 ways. The letters P, L, A, T, N create 6 gaps, in which O’s are arranged. ∴ Two O’s in 6 gaps can be arranged in ways = ways = ways = 3 × 5 ways = 15 ways ∴ Total number of arrangements if the two O’s are never together = 120 × 15 = 1800
(ii) When consonants and vowels occupy alternate positions:
There are 4 consonants and 3 vowels in the word PLATOON.
∴ At odd places consonants occur and at even places vowels occur.
4 consonants can be arranged among themselves in 4! ways
3 vowels in which O occurs twice and A occurs once.
∴ They can be arranged in ways
∴ Required number of arrangements if the consonants and vowels occupy alternate positions = 4! ×
= 4 × 3 × 2 ×
= 72