Maharashtra State Board 11th Maths Solutions Chapter 3 Permutations and Combination Ex 3.4
Solution & Step-by-Step Answer:
(i) There are 5 distinct letters in the word DIVYA. ∴ Number of permutations of the letters of the word DIVYA = 5! = 120
(ii) There are 9 letters in the word SHANTARAM in which ‘A’ is repeated 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’ is repeated 3 times and ‘R’ is repeated 2 times.
∴ Number of permutations of the letters of the word REPRESENT =
=
= 30240
(iv) There are 7 distinct letters in the word COMBINE.
∴ Number of permutations of the letters of the word COMBINE = 7! = 5040
(v) There are 10 letters in the word BALBHARATI in which ‘B’ is repeated 2 times and ‘A’ is repeated 3 times.
∴ Number of permutations of the letters of the word BALBHARATI =
=
= 302400
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 possible = = = 9 × 4 × 7 × 5 = 1260
Solution & Step-by-Step Answer:
A coin is tossed 8 times. All heads are identical and all tails are identical. (a) 4 heads and 4 tails are to be obtained. ∴ Number of ways it can be obtained = = = 70
(b) 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 it can be obtained =
= + 8 + 1
= 28 + 8 + 1
= 37
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’ is repeated 2 times, ‘A’ repeated 3 times and ‘T’ repeated 2 times. ∴ Required number of arrangements = When all the vowels, i.e., ‘A’, ‘A’, ‘A’, ‘E’, ‘I’ are to be kept together. Let us consider them as one unit. Number of arrangements of these vowels among themselves = ways. This unit is to be arranged with 7 other letters in which ‘M’ and ‘T’ repeated 2 times each. ∴ Number of such arrangements = ∴ Required 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 words can be formed =
(a) When letters R and H are never together.
Other than 2R, 2H there are 4A, 1S, 1T, 1M.
These letters can be arranged in ways = 210.
These seven letters create 8 gaps in which 2R, 2H are to be arranged.
Number of ways to do = = 420
Required number of arrangements = 210 × 420 = 88200.
(b) 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, which can be arranged among themselves in = 1 way.
This unit is to be arranged with 7 other letters in which ‘H’ is repeated 2 times, ‘R’ is repeated 2 times.
∴ Total number of arrangements =
Solution & Step-by-Step Answer:
To find the number of different words when ‘R’ is taken thrice, ‘S’ is taken twice and ‘T’ is taken twice. ∴ Total number of letters available = 7, of which ‘S’ and ‘T’ repeat 2 times each, ‘R’ repeats 3 times. ∴ Required number of words = = = 7 × 6 × 5 = 210
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 the other 4 letters in which ‘M’ repeats twice is to be arranged. ∴ Required number of arrangements = = = 60
Solution & Step-by-Step Answer:
There are 12 letters in the word CONSTITUTION, in which ‘O’, ‘N’, ‘I’ repeat two times each, ‘T’ repeats 3 times. When the arrangement starts and ends with ‘N’, other 10 letters can be arranged between two N, in which ‘O’ and ‘I’ repeat twice each and ‘T’ repeats 3 times. ∴ Required number of arrangements =
Solution & Step-by-Step Answer:
There are 7 letters in the word ARRANGE in which ‘A’ and ‘R’ repeat 2 times each. ∴ Number of ways to arrange the letters of word ARRANGE = = 1260 Consider the words in which 2A are together and 2R are together. Let us consider 2A as one unit and 2R as one unit. These two units with remaining 3 letters can be arranged in = = 30 ways. Number of arrangements in which neither 2A together nor 2R are together = 1260 – 30 = 1230
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 = = 5! = 120 Case II: Numbers are formed from 2, 2, 3, 3 and any one of 4 or 5 Number of such numbers = = 60 Required number of numbers = 120 + 60 = 180
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! = 24 ways. There are 3 even digits, i.e. 4, 6, 8. ∴ They can be arranged at 3 even places among themselves in 3! = 6 ways. ∴ Required number of numbers formed = 24 × 6 = 144
Solution & Step-by-Step Answer:
A 6-digit number is to be formed using digits of 659942, in which 9 repeats twice. ∴ Required number of numbers formed = = = 360 A 6-digit number is to be formed using the same digits that are divisible by 4. For a number to be divisible by 4, the last two digits should be divisible by 4, i.e. 24, 52, 56, 64, 92 or 96. Case I: When the last two digits are 24, 52, 56 or 64. As the digit 9 repeats twice in the remaining four numbers, the number of arrangements = = 12 ∴ 6-digit numbers that are divisible by 4 so formed are 12 + 12 + 12 + 12 = 48. Case II: When the last two digits are 92 or 96. As each of the remaining four numbers are distinct, the number of arrangements = 4! = 24 ∴ 6-digit numbers that are divisible by 4 so formed are 24 + 24 = 48. ∴ Required number of numbers framed = 48 + 48 = 96
Solution & Step-by-Step Answer:
There are 6 letters in the word INDIAN in which I and N are repeated twice. Number of different words that can be formed using the letters of the word INDIAN = = = 180 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. ∴ 2 N can be arranged in = 1 way. ∴ Required number of words = 60 × 1 = 60
Solution & Step-by-Step Answer:
There are 7 letters in the words PLATOON in which ‘O’ repeat 2 times. (a) 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 can take their places in 6P2 ways. But ‘O’ repeats 2 times. ∴ Two O’s can be arranged in = = = 3 × 5 = 15 ways ∴ Required number of arrangements = 120 × 15 = 1800
(b) 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.
Now, vowels and consonants should occupy alternate positions.
∴ Required number of arrangements = 4! ×
= 4 × 3 × 2 ×
= 72