Maharashtra State Board 11th Maths Solutions Chapter 3 Permutations and Combination Ex 3.3
Solution & Step-by-Step Answer:
nP6 : nP3 = 120 : 1 ∴ (n – 3) (n – 4) (n – 5) = 120 ∴ (n – 3) (n – 4) (n – 5) = 6 × 5 × 4 Comparing on both sides, we get n – 3 = 6 ∴ n = 9
Solution & Step-by-Step Answer:
(m+n)P2 = 56 (m + n) (m + n – 1) = 56 Let m + n = t t(t – 1) = 56 t2 – t – 56 = 0 (t – 8) (t + 7) = 0 t = 8 or t = -7 m + n = 8 or m + n = -7 But m + n ≠ -7 ∴ m + n = 8 ……(i) Also, (m-n)P2 = 12 (m – n) (m – n – 1) = 12 Let m – n = a a(a – 1) = 12 a2 – a – 12 = 0 (a – 4)(a + 3) = 0 a = 4 or a = -3 m – n = 4 or m – n = -3 But m – n ≠ -3 ∴ m – n = 4 ……(ii) Adding (i) and (ii), we get 2m = 12 ∴ m = 6 Substituting m = 6 in (ii), we get 6 – n = 4 ∴ n = 2
Solution & Step-by-Step Answer:
(14 – r)(13 – r) = 8 × 7 Comparing on both sides, we get 14 – r = 8 ∴ r = 6
Solution & Step-by-Step Answer:
Solution & Step-by-Step Answer:
There are 7 letters in the word MADHURI. (a) A 4 letter word is to be formed from the letters of the word MADHURI and repetition of letters is allowed. ∴ 1st letter can be filled in 7 ways. 2nd letter can be filled in 7 ways. 3rd letter can be filled in 7 ways. 4th letter can be filled in 7 ways. ∴ Total no. of ways a 4-letter word can be formed = 7 × 7 × 7 × 7 = 2401
(b) When repetition of letters is not allowed, the number of 4-letter words formed from the letters of the word MADHURI is7P4= = 840
Solution & Step-by-Step Answer:
There are 9 letters in the word ALGORITHM. (a) When vowels are always together. There are 3 vowels in the word ALGORITHM (i.e., A, I, O). Let us consider these 3 vowels as one unit. This unit with 6 other letters is to be arranged. ∴ The number of arrangement = 7P7 = 7! = 5040 3 vowels can be arranged among themselves in 3P3 = 3! = 6 ways. ∴ Required number of arrangements = 7! × 3! = 5040 × 6 = 30240
(b) When no two vowels are together.
There are 6 consonants in the word ALGORITHM,
they can be arranged among themselves in6P6= 6! = 720 ways.
Let consonants be denoted by C.
_C _C_ C _C_C_C
There are 7 places marked by ‘_’ in which 3 vowels can be arranged.
∴ Vowels can be arranged in7P3= = 210 ways.
∴ Required number of arrangements = 720 × 210 = 151200
(c) When consonants are at even positions.
There are 4 even places and 6 consonants in the word ALGORITHM.
∴ 6 consonants can be arranged at 4 even positions in 6P4 = = 360 ways.
Remaining 5 letters (3 vowels and 2 consonants) can be arranged in odd position in5P5= 5! = 120 ways.
∴ Required number of arrangements = 360 × 120 = 43200
(d) When O is the first and T is the last letter.
All the letters of the word ALGORITHM are to be arranged among themselves such that arrangement begins with O and ends with T.
∴ Position of O and T are fixed.
∴ Other 7 letters can be arranged between O and T among themselves in7P7= 7! = 5040 ways.
∴ Required number of arrangements = 5040
Solution & Step-by-Step Answer:
In 1st row, 6 teachers can be arranged among themselves in 6P6 = 6! ways. In the 2nd row, 12 boys can be arranged among themselves in 12P12 = 12! ways. No two girls are together. So, there are 13 places formed by 12 boys in which 6 girls occupy any 6 places in 13P6 ways. ∴ Required number of arrangements
Solution & Step-by-Step Answer:
There are 7 letters in the word HISTORY (a) When ‘Y’ and ‘T’ are together. Let us consider ‘Y’ and ‘T’ as one unit. This unit with other 5 letters are to be arranged. ∴ The number of arrangement of one unit and 5 letters = 6P6 = 6! = 720 Also, ‘Y’ and ‘T’ can be arranged among themselves in 2P2 = 2! = 2 ways. ∴ A total number of arrangements when Y and T are always together = 6! × 2! = 120 × 2 = 1440
(b) When ‘Y’ is next to ‘T’.
Let us take this (‘Y’ next to ‘T’) as one unit.
This unit with 5 other letters is to be arranged.
∴ The number of arrangements of 5 letters and one unit =6P6= 6! = 720
Also, ‘Y’ has to be always next to ‘T’.
∴ They can be arranged among themselves in 1 way only.
∴ Total number of arrangements possible when Y is next to T = 720 × 1 = 720
(c) When there is no restriction.
7 letters can be arranged among themselves in7P7= 7! ways.
∴ The total number of arrangements possible if there is no restriction = 7!
(d) When begin and end with a vowel.
There are 2 vowels in the word HISTORY.
All other letters of the word HISTORY are to be arranged between 2 vowels such that the arrangement begins and ends with a vowel.
The other 5 letters can be filled between the two vowels in5P5= 5! = 120 ways.
Also, 2 vowels can be arranged among themselves at first and last places in2P2= 2! = 2 ways.
∴ Total number of arrangements when the word begins and ends with vowel = 120 × 2 = 240
(e) When a word ends in ST.
As the arrangement ends with ST,
the remaining 5 letters can be arranged among themselves in5P5= 5! = 120 ways.
∴ Total number of arrangements when the word ends with ST = 120
(f) When a word begins with S and ends with T.
As arrangement begins with S and ends with T,
the remaining 5 letters can be arranged between S and T among themselves in5P5= 5! = 120 ways.
Total number of arrangements when the word begins with S and ends with T = 120
Solution & Step-by-Step Answer:
There are 4 consonants S, L, P, R, and 3 vowels A, O, U in the word SOLAPUR. Consonants and vowels are to be alternated. ∴ Vowels must occur in even places and consonants in odd places. ∴ 3 vowels can be arranged at 3 even places in 3P3 = 3! = 6 ways. Also, 4 consonants can be arranged at 4 odd places in 4P4 = 4! = 24 ways. Required number of arrangements = 6 × 24 = 144
Solution & Step-by-Step Answer:
(a) A 4 digit number is to be made from the digits 1, 2, 4, 5, 6, 8 such that digits can be repeated. ∴ Unit’s place digit can be filled in 6 ways. 10’s place digit can be filled in 6 ways. 100’s place digit can be filled in 6 ways. 1000’s place digit can be filled in 6 ways. ∴ Total number of numbers that can be formed = 6 × 6 × 6 × 6 = 1296
(b) A 4 different digit number is to be made from the digits 1, 2, 4, 5, 6, 8 without repetition of digits.
∴ 4 different digits are to be arranged from 6 given digits which can be done in6P4ways.
∴ Total number of numbers that can be formed
=
=
= 360
Solution & Step-by-Step Answer:
A number between 100 and 1000 that can be formed from the digits 0, 1, 2, 3, 4, 5 is of 3 digits, and repetition of digits is not allowed. ∴ 100’s place can be filled in 5 ways as it is a non-zero number. 10’s place digits can be filled in 5 ways. Unit’s place digit can be filled in 4 ways. ∴ Total number of ways the number can be formed = 5 × 5 × 4 = 100
Solution & Step-by-Step Answer:
A number of 6 different digits is to be formed from the digits 3, 4, 5, 6, 7, 8 which can be done in 6P6 = 6! = 720 ways. (a) If the number is to be divisible by 5, the unit’s place digit can be 5 only. ∴ it can be arranged in 1 way only. The other 5 digits can be arranged among themselves in 5P5 = 5! = 120 ways. ∴ Required number of numbers divisible by 5 = 1 × 120 = 120
(b) If the number is not divisible by 5,
unit’s place can be any digit from 3, 4, 6, 7, 8.
∴ it can be arranged in 5 ways.
Other 5 digits can be arranged in5P5= 5! = 120 ways.
∴ Required number of numbers not divisible by 5 = 5 × 120 = 600
Solution & Step-by-Step Answer:
There is a total of 26 alphabets. A code word contains 2 English alphabets. ∴ 2 alphabets can be filled in 26P2 = = = 650 ways. Also, alphabets to be followed by two distinct non-zero digits from 1 to 9 which can be filled in 9P2 = = 72 ways. ∴ Total number of a code words = 650 × 72 = 46800. To find the number of codewords end with an even integer. 2 alphabets can be filled in 650 ways. The digit in the unit’s place should be an even number between 1 to 9, which can be filled in 4 ways. Also, 10’s place can be filled in 8 ways. ∴ Total number of codewords = 650 × 4 × 8 = 20800
Solution & Step-by-Step Answer:
There are 5 letters and 3 post boxes and any number of letters can be posted in all three post boxes. ∴ Each letter can be posted in 3 ways. ∴ Total number of ways 5 letters can be posted = 3 × 3 × 3 × 3 × 3 = 243
Solution & Step-by-Step Answer:
There are 11 distinct objects and 4 are to be taken at a time. (a) The number of permutations of n distinct objects, taken r at a time, when one particular object will always occur is ∴ In 2880 permutations of 11 distinct objects, taken 4 at a time, one particular object will always occur.
(b) When one particular object will not occur, then 4 objects are to be arranged from 10 objects which can be done in10P4= 10 × 9 × 8 × 7 = 5040 ways.
∴ In 5040 permutations of 11 distinct objects, taken 4 at a time, one particular object will never occur.
Solution & Step-by-Step Answer:
(i) 5 books arranged in 5P5 = 5! = 120 ways.
(ii) 2 books are together.
Let us consider two books as one unit. This unit with the other 3 books can be arranged in4P4= 4! = 24 ways.
Also, two books can be arranged among themselves in2P2= 2 ways.
∴ Required number of arrangements = 24 × 2 = 48
(iii) Say books are B1, B2, B3, B4, B5are to be arranged with B1, B2never together.
B3, B4, B5can be arranged among themselves in3P3= 3! = 6 ways.
B3, B4, B5create 4 gaps in which B1, B2are arranged in4P2= 4 × 3 = 12 ways.
∴ Required number of arrangements = 6 × 12 = 72
Solution & Step-by-Step Answer:
3 boys and 3 girls are to be arranged in a row. (i) When there are no restrictions. ∴ Required number of arrangements = 6! = 720
(ii) When there is a girl at each end.
3 girls can be arranged at two ends in
3P2= = 3 × 2 = 6 ways.
And remaining 1 girl and 3 boys can be arranged between the two girls in4P4= 4! = 24 ways.
∴ Required number of arrangements = 6 × 24 = 144
(iii) Boys and girls are at alternate places.
We can first arrange 3 girls among themselves in3P3= 3! = 6 ways.
Let girls be denoted by G.
G – G – G –
There are 3 places marked by ‘-’ where 3 boys can be arranged in 3! = 6 ways.
∴ Total number of such arrangements = 6 × 6 = 36
OR
Similarly, we can first arrange 3 boys in 3! = 6 ways
and then arrange 3 girls alternately in 3! = 6 ways.
∴ Total number of such arrangements = 6 × 6 = 36
∴ Required number of arrangements = 36 + 36 = 72
(iv) All boys sit together.
Let us consider all boys as one group.
This one group with the other 3 girls can be arranged4P4= 4! = 24 ways.
Also, 3 boys can be arranged among themselves in3P3= 3! = 6 ways.
∴ Required number of arrangements = 24 × 6 = 144