OP Malhotra Class 11 Maths Solutions Chapter 12 Permutations and Combinations Ex 12(c)

Continuous practice using OP Malhotra Maths Class 12 Solutions Chapter 12 Permutations and Combinations Ex 12(c) can lead to a stronger grasp of mathematical concepts.

S Chand Class 11 ICSE Maths Solutions Chapter 12 Permutations and Combinations Ex 12(c)

Question 1.
Of 12 different books a shelf will hold five ; how many different arrangements may be made on the shelf ?
Solution:
total no. of books =12
since it is given that a shelf holds five books
Thus required no. of different arrangement may be made on shelf = ¹²P₅ = \(\frac{12 !}{(12-5) !}\)
= \(\frac{12 !}{7 !}\) = 12 × 11 × 10 × 9 × 8 = 95040

Question 2.
In how many ways can be letters of the following words be arranged :
(i) RADIO
(ii) FOREIGN ?
Solution:
(i) The required no. of ways in which five letters of the word RADIO can be arranged = 5! = 120
(ii) The required no. of ways in which seven letters of the word FOREIGN can be arranged = 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040

Question 3.
In how many other ways can the letters of the word ‘SIMPLETON’ be arranged ?
Solution:
Total no. of letters in word SIMPLETON = 9
∴ required no. of ways in which letters of the given word can be arranged = 9! – 1
= 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 -1 = 362,880 – 1 = 362779

Question 4.
How many different words beginning and ending with a constant can be made out of the letters of the word ‘EQUATION’ ?
Solution:
In word EQUATION,
no. of vowels = 5 {E, Q, A, I, 0} and no. of consonants = 3 {Q, T, N}
For beginning and ending of words i.e. for 2 positions,
we have 3 consonants and this can be done in 3P2 ways.
The remaining 6 letters can be arranged in 6! different ways
∴ required no. of such words = ³P₂ × 6! = \(\frac{3 !}{(3-2) !}\) × 6! = 6 × 6 × 5 × 4 × 3 × 2 × 1 = 4320

Question 5.
How many permutations can be made out of the letters of the word ‘TRIANGLE’ ? How many of these will begin with T and end with E ?
Solution:
Total no. of letters in word TRIANGLE = 8
∴ required no. of permutations can be made out of 8 letters of given word = 8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 40320
Since we want to find permutation which begin with T and end with E so beginning and ending are fixed and arranged in \(\lfloor 1\) way each. The remaining six letters can be arranged among themselves in 6 places in ⁶P₆ ways.
∴ required no. of such permutations = \(\lfloor 1\) × \(\lfloor 1\) × ⁶P₆ = 6! = 720

Question 6.
How many different words can be formed of the letters of the word ‘MALENKOV’ so that
(i) no two vowels are together,
(ii) the vowels may occupy odd places ?
Solution:
In given word MALENKOV, no. of vowels = 3 {A, E, 0} and no. of consonants = 5
(i) Now we want to find words in which no two vowels are together.
First of all, we fix the position of five consonants and this can be done in 5! ways.
Now we place 3 vowels × C₅ × C₄ × C₃ × C₂ × C₁ × at 6 positions denoted by ‘x’ sign.
This can be done in ⁶P₆
∴ required no. of such different words in which no two vowels come together
= 5! × ⁶P₃ = 120 × \(\frac{6 !}{3 !}\) = 120 × 120 = 14400

(ii) Since there are 8 letters in which 4 odd places.
firstly we arrange 3 vowels at 4 places and this can be done in ⁴P₃ ways and the remaining 5 letters arranged in 5 places in ⁵P₅ way
∴ required no. of such words = ⁴P₃ × ⁵P₅ = \(\frac{4 !}{(4-3) !}\) × \(\frac{5 !}{(5-5) !}\) = 4! × 5! = 24 × 120 = 2880

Question 7.
In how may ways can be letters of the word ‘COMBINE’ be arranged so that;
(i) the vowels are never separated;
(ii) all the vowels nevercome together;
(iii) vowels occupy only the odd places ?
Solution:
In word COMBINE
Total no. of vowels = 3{O, I, E}
Total no. of consonants = 4{C, M, B, N}

(i) We want to find no. of words in which vowels are never separated i.e. they occur together so they form one group and remaining 4 consonants i.e. 5 letters can be arranged in 5 ! ways.
The group containing 3 vowels can be arranged themselves in 3 ! ways
∴ required no. of words = 5 ! × 3 ! = 120 × 6 = 720

(ii) Total no. of ways in which 7 letters of given word can be arranged = 7 !
∴ required no. of ways in which all the vowels never come together
= Total no. of words – Total no. of words in which vowels occur together
= 7! – 720 = 5040 – 720 = 4320

(iii) Firstly we arrange 3 vowels at four odd positions in ⁴P₃ ways and remaining 4 consonants can be arranged in 4! ways.
∴ required no. of such words = ⁴P₃ × 4! = 4! × 4! = 576

Question 8.
Three persons have 4 coats, 5 waistcoats, and 6 hats. Find in how many ways can they put on the clothes.
Solution:
Regarding coats, first person have 4 choices, second person have 3 choices and third person have 2 choices.
∴ Total no. of ways in which 3 persons can put coat = 4 × 3 × 2 = 24
For 3 persons we have 5 waistcoats and this can be done in ⁵P₃ ways
For 3 persons, we have 6 hats, and this can be done in ⁶P₃ ways
Hence the required no. of ways in which they can put on the clothes = ⁴P₃ × ⁵P₃ × ⁶P₃
= 4! × \(\frac{5 !}{2 !}\) × \(\frac{6 !}{3 !}\) = 24 × 5 × 4 × 3 × 6 × 5 × 4 = 24 × 60 × 20 = 172800

Question 9.
If out of 6 flags any number of flags can be shown at a time, find how many different signals can be made out of them.
Solution:
Given total no. of flags = 6
∴ no. of different signals made by using 1 flag = ⁶P₁
no. of different signals made by using 2 flags = ⁶P₂
no. of different signals made by using 3 flags = ⁶P₃
no. of different signals made by using 4 flags = ⁶P₄
no. of different signals made by using 5 flags = ⁶P₅
no. of different signals made by using 6 flags = ⁶P₆
Thus required total no. of signals can be made using all flags
= ⁶P₁ + ⁶P₂ + ⁶P₃ + ⁶P₄ + ⁶P₅ + ⁶P₆
= \(\frac{6 !}{5 !}\) + \(\frac{6 !}{4 !}\) + \(\frac{6 !}{3 !}\) + \(\frac{6 !}{2 !}\) + \(\frac{6 !}{1 !}\) + \(\frac{6 !}{0 !}\)
=6 + 6 × 5 + 6 × 5 × 4 + 6 × 5 × 4 × 3 + 6 × 5 × 4 × 3 × 2 + 61
= 6 + 30 + 120 + 360 + 720 + 720 = 1956

Question 10.
In how many ways can 9 things be arranged taken 4 at a time, and in how many of these arrangements will a particular thing be included ?
Solution:
The required no. of ways in which 9 things can be arranged taken 4 at a time = ⁹P₄
= \(\frac{9 !}{(9-4) !}\) = \(\frac{9 !}{5 !}\) = 9 × 8 × 7 × 6 = 3024
∴ required no. of arrangements in which a particular thing be included
= ⁸P₃ × 4 = \(\frac{9 !}{5 !}\) = 8 × 7 × 6 × 4 = 1344

Question 11.
How many different numbers of 4 digits each can be formed with the ten digits 0,1,2,… 9, when digits are not repeated ?
Solution:
Given 10 digits are 0, 1, 2 ….. 9.
Here thousand place can be filled in 9 ways (excluding 0).
Hundred place can be filled by 9 ways (including 0) since digits are not repeated.
Thus tens place can be filled in 8 ways and unit place can be filled in 7 ways

∴ required no. of different 4 digit numbers = 9 × 9 × 8 × 7 = 4536

Question 12.
From the digits 1, 2, 3, 4, 5, 6, how many three-digit odd numbers can be formed when the repetition of the digits is not allowed.
Solution:
We know that every odd number is having odd digit at unit place.
Thus unit place can be filled by 3 digits i.e. 1, 3 and 5

Ten place can be filled in 5 different ways since repetition of digits is not allowed.
Therefore, hundred place can be filled in 4 ways
∴ required no. of three digit odd numbers = 4 × 5 × 3 = 60

Question 13.
(i) How many different numbers of six digits can be formed with the digits 3, 1, 7, 0,9, 5 ?
(ii) How many of them are divisible by 10 ?
(iii) How many of them will have zero in the ten’s place ?
Solution:
(i) Given digits are 3, 1, 7, 0, 9, 5 and repetitions of digits is not allowed.
sixth place from left can be filled by 5 digits (excluding 0).
The remaining five places can be filled by 5 digits in ⁵P₅ ways.
required no. of different six digit numbers = 5 × ⁵P₅ = 5 × 5! = 5 × 120 = 600

(ii) We know that a number is divisible by 10 if its unit place contain 0.
Thus unit place filled by 1 way only and remaining five places filled by five digits in ⁵P₅ ways.
∴ required no. of such numbers = 1 × ⁵P₅ = 1 × 5! = 120

(iii) Now we want to find numbers having 0 in tens place i.e. tens place filled by 1 digit i.e. 0 in 1 way only. The remaining five places can be filled by five digits in ⁵P₅ ways.
Thus required no. of such numbers =1 × ⁵P₅ = 1 × 5! = 120

Question 14.
How many 5-digit telephone numbers can be formed with the digits 0, 1, 2,…, 8, 9 if each number starts with 35 and no digit appears more than once ?
Solution:
Given no. of digits = 10
we want to find 5-digit telephone numbers with these given digits in which no digit repeated and each number starts with 35.
i.e. first two places are fixed and for remaining 3 places, we have 8 digits and this can be done in ⁸P₃ ways.
Thus required no. of such 5 digit telephone numbers = 1 × ⁸P₃ = \(\frac{8 !}{5 !}\) = 8 × 7 × 6 = 336

Question 15.
There are 5 boys and 3 girls. In how many ways can they stand in a row so that no two girls are together ?
Solution:
Firstly, we fix the position of 5 boys and this can be done in 5! ways.
Now we want to find the no. of ways in which no two girls are together. This is only possible when girls and sitting on position shown by x mark.
Thus 3 girls placed on 6 places in ⁶P₃ ways

∴ required no. of ways = 5! × ⁶P₃ =120 × 6 × 5 × 4 = 14400

Question 16.
There are 5 red, 4 white and 3 blue marbles in a bag. They are drawn one by one and arranged in a row. Assuming that all the 12 marbles are drawn, determine the number of different arrangements.
Solution:
given total no. of marbles = 12
no. of red marbles = 5
no. of white marbles = 4
and no. of blue marbles = 3
since marbles are drawn one by one and arranged in a row.
Thus required no. of diff. arrangements

Question 17.
How many 7-digit numbers can be formed using the digits 1, 2, 0, 2, 4, 2 and 4 ?
Solution:
Given digits are 1, 2, 0, 2, 4, 2 and 4
Total no. of different arrangements = \(\frac{7 !}{3 ! 2 !}\)
Total no. of different arrangement having 0 in extreme position = \(\frac{6 !}{3 ! 2 !}\)
∴ required no. of 7 digit numbers = \(\frac{7 !}{3 ! 2 !}\) – \(\frac{6 !}{3 ! 2 !}\) = \(\frac{7 !-6 !}{3 ! \times 2 !}\) = \(\frac{(7-1) 6 !}{6 \times 2}\) = \(\frac{6 \times 6 \times 5 \times 4 \times 3 \times 2}{12}\) = 360

Question 18.
(i) How many different words can be formed with the letters of the word ‘BHARAT’ ?
(ii) In how many of these B and H are never together ?
(iii) How many of these begin with B and end w ith T ?
Solution:
(i) In word B, H, A, R, A, T, there are two A’s, 1B, 1R, 1H and 1T.
∴ required no. of different words = \(\frac{6 !}{2 ! 1 ! 1 ! 1 ! 1 !}\) = \(\frac{720}{2}\) = 360

(ii) Firstly, we want to find the no. of words, in which B and H are together i.e. B and H forms one group. Thus one group and remaining four letters can be arranged in \(\frac{5 !}{2 ! 1 ! 1 !}\) i.e. \(\frac{5 \times 4 \times 3 \times 2}{2}\)
i.e. 60 ways and group itself can be arranged in 2 ways.
∴ Total no. of words in which B and H are together = 60 × 2 = 120
Hence the required no. of different words in which B and H are never together
= Total no. of words – Total no. of vowels in which B and H come together = 360 – 120 = 240

(iii) We want to find the no. of words that begin with B and end with T.
So starting and end position are fixed and remaining four positions are filled by remaining four letters and this can be done in \(\frac{4 !}{2 ! 1 ! 1 !}\) i.e. 12 ways
∴ required no. of words = 12

Question 19.
(i) Find how many arrangements can be made with the letters of the word ‘MATHEMATICS’?
(ii) In how many of them the vowels occur together ?
Solution:
(i) In word MATHEMATICS, we have 2M’s, 2A’s, 2T’s, 1H, 1E, 1C, 1I and 1S.
∴ required no. of arrangements = \(\frac{11 !}{2 ! 2 ! 2 !}\)
= \(\frac{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2}{2 \times 2 \times 2}\) = 4989600

(ii) We want to find no. of words in which vowels occur together.
Here, first of all, we take all vowels A, A, I, E as one group. This group and remaining 7 letters i.e. in total 8 letters can be arranged themselves in \(\frac{8 !}{2 ! 2 !}\) ways. Further, vowels can be arranged themselves in \(\frac{4 !}{2 ! 2 !}\) ways.
Thus, required no. of arrangements = \(\frac{8 !}{2 ! 2 !}\) × \(\frac{4 !}{2 !}\) = \(\frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2}{2 \times 2 \times 2}\) × 24 = 120960

Question 20.
Ten different books are arranged on a shelf. Find the number of different ways in which this can be done, if two specified books are (a) to be together, (b) not to be together.
Solution:
(a) Two specific books together forms a group and arranged themselves in 2! ways. This group and remaining 8 books i.e. 9 books can be arranged themselves in 9! ways.
∴ required no. of different ways = 2! × 9! = 2 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 = 725760

(b) Required no. of ways in which two specific books not together
= Total no. of ways – Total no. of ways in which two specific books together
= 10! – 2! × 9! = 9! (10 – 2) = 8 × 9! = 2903040

Question 21.
In how many ways can 20 books be arranged on a shelf so that a particular pair of books shall not come together ?
Solution:
Total no. of ways in which 20 books can be arranged on a shelf = 20!
when a particular pair of books come together they form a group. Now this group and remaining 18 books i.e. in total 19 books can be arranged in 19! ways.
Further, pair of books can be arranged themselves in 2 ways.
Thus required no. of ways in which a particular pair of books shall not come together = Total no. of arrangements of books on shelf – Total no. of arrangements in which pair of books come together
= 20! – 19! 2! = 19! (20 – 2) = 18 × 19!

Question 22.
Find the number of permutations of the letters of the words
(i) INDIA
(ii) ALLHABAD
(iii) CHANDIGARH
(iv) COMMISSION.
Solution:
(i) Total no. of permutations of the letters of the word INDIA = \(\frac{5 !}{2 ! 1 ! 1 ! 1 !}\) = \(\frac{5 \times 4 \times 3 \times 2 \times 1}{2}\) = 60

(ii) Total no. of permutations of the letters of the word ALLAHABAD = \(\frac{9 !}{4 ! 2 ! 1 ! 1 ! 1 !}\) [Here we have 4A’s and 2L’s]
= \(\frac{9 \times 8 \times 7 \times 6 \times 5}{2}\) = 7560

(iii) In word CHANDIGARH, there are 2A’s, 1C, 1H, 1N, 1D, 1I, 1Q 1R and 1H.
∴ required no. of permutations = \(\frac{10 !}{2 !}\) = \(\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2}{2}\) = 907200

(iv) In word COMMISSION, we have 2M’s, 2S’s, 2O’s, 2I’s, 1C and IN
∴ required no. of permutations = \(\frac{10 !}{2 ! 2 ! 2 ! 2 !}\) = \(\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2}{2 \times 2 \times 2 \times 2}\) = 226800

Question 23.
Find the number of ways in which five identical balls can be distributed among ten identical boxes, if not more than one can go into a box.
Solution:
Since more than one ball cannot go into one box.
∴ Total no. of ways in which five identical balls can be distributed among ten identical boxes = \(\frac{10 !}{5 ! 5 !}\) = \(\frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2}\) = 252

Question 24.
How many numbers are there in all which consist of 5 digits ?
Solution:
Since 5th place from left can be filled by 9 digits except 0 i.e. in 9 different ways.
The remaining four places can be filled by 10 ways each.
Thus required no. of 5-digit numbers = 9 × 10 × 10 × 10 × 10 = 90000

Question 25.
In how many ways can 5 prizes be distributed among 4 students, when each student may receive any number of prizes ?
Solution:
For first student, we have five prizes. Since each student may receive any no. of prizes.
Thus for 2nd, 3rd and 4th student we have 5 choices each.
Hence required no. of ways in which 5 prizes be distributed among 4 students when each student may receive any no. of prizes = 5 × 5 × 5 × 5 = 625

Question 26.
In how many ways can 4 letters be posted in four letter boxes in a village ? If all the three letters are not posted in the same letter box, find the corresponding number of ways of posting.
Solution:
For first letter, we have 4 choices i.e. letter can be posted to any of the four letter boxes.
Similarly for 2nd and 3rd letter we have 4 choices each.
Thus required no. of such ways = 4 × 4 × 4 = 64
when all the three letters are not posted in same letter box so in total, there are 4 ways in which all the three letters are posted in same letter box.
∴ required no. of ways in which all the three letters are not posted in same letter box = 64 – 4 = 60

Question 27.
In how many ways can 8 people sit around a table ?
Solution:
∴ required no. of ways in which 8 people sit around a table = \(\lfloor n-1\) (Here n = 8)
= \(\lfloor 8-1\) = \(\lfloor 7\) = 7 × 6 × 5 × 4 × 3 × 2 = 5040

Question 28.
(i) In how many ways can 10 people sit around a table so that all shall not have the same neighbours in any two arrangements ?
(ii) In how many ways can 20 persons be seated round a table if there are 9 chairs.
Solution:
(i) 10 persons can sit around a table in (10 – 1)! ways i.e. 9! ways but each person will have the same neighbours in clockwise and anti-clockwise arrangements.
∴ required no. of arrangements = \(\frac{1}{2}\) × 9!

(ii) We know that, no. of circular permutations of n different things taken r at a time when clockwise
and anticlockwise arrangements are taken as different is \(\frac{{ }^n \mathrm{P}_r}{r}\)
∴ required no. of ways in which 20 persons be seated round a table if there are 9 chairs = \(\frac{{ }^{20} \mathrm{P}_9}{9}\)

Question 29.
A committee of 11 members sits at a round table. In how many ways can they be seated if the ‘President’ and the ‘Secretary’ choose to sit together ?
Solution:
Since it is given that a committee of 11 members sits at a round table. Since President and Secretary choose to sit together so they form a group and can be arranged in 2 ways. This group and remaining 9 members sits round a table in (10 – 1)! ways i.e. 9! ways.
Thus required no. of ways = 2 × 9! = 2 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 = 725750

Question 30.
In how many ways can 30 different pearls be arranged to form a necklace ?
Solution:
Required no. of ways in which 30 different pearls be arranged to form a necklace

Question 31.
In how many ways 6 gentlemen and 3 ladies can be seated round a table so that every gentleman may have a lady by his side.
Solution:
6 gentlemen can be seated round a table in (6 – 1)! ways i.e. 5! The seating arrangement of the ladies be shown in figure given alongside.
Thus 3 ladies can be arranged in 3! ways.
∴ Total no. of arrangements = 5! × 3!

If the arrangements are made in opposite direction, then the
no. of arrangements = \(\lfloor 5\) × \(\lfloor 3\)
∴ required no. of arrangements = \(\lfloor 5\) × \(\lfloor 3\) + \(\lfloor 5\) × \(\lfloor 3\) = 2 × 120 × 6 = 1440

Question 32.
The letters of the word ZENITH are written in all possible orders. How many words are possible if all these words are written out as in a dictionary ? What is the rank of the word ZENITH ?
Solution:
Since the words in dictionary are arranged in alphabetical order.
(i) Starting with E, The remaining letters H, I, N, T, Z, can be arranged in 5! = 120 ways. Thus, there are 120 words starting with E.
(ii) Starting with H, the remaining letters A, I, N, T, Z can be arranged in 5! = 120 ways.
(iii) Starting with I, the remaining letters A, H, N, T, Z can be arranged in \(\lfloor 5\) = 120 ways.
(iv) Starting with N, the remaining letters A, H, I, T, Z can be arranged in \(\lfloor 5\) = 120 ways.
(v) Starting with T, the remaining letters A, H, I, N, Z can be arranged in \(\lfloor 5\) = 120 ways.
So we have 120 + 120 + 120 + 120 + 120 = 600 word constructed so far.
(vi) Starting with Z, the remaining letters A, H, I, N, T can be arranged in \(\lfloor 5\) = 120 ways.
Thus no. of words starting with Z is also 120 out of these words, one word is ZENITH.
No. of words starting with ZEH = 3! = 6
No. of words starting with ZEI = 3! = 6
Thus so far, we have constructed 120 + 120 + 120 + 120 + 120 + 6 + 6 = 612
Now no. of words starting with ZEN be also 6 out of which one word is ZENITH.
The first words starting with ZEN be ZENHIT which is 613th word.
∴ 614th word be ZENHTI, 615 th word be ZENIHT and 616 th word be ZENITH