Permutation and Combination Solution

1. How many words with or without meaning, can be formed by using all the letters of the word, 'DELHI' using each letter exactly once?A. 720 B. 24 C. None of these D. 120

Sol: The word 'DELHI' has 5 letters and all these letters are different.Total words (with or without meaning) formed by using all these 5 letters using each letter exactly once = Number of arrangements of 5 letters taken all at a time= 5P5 = 5! = 5 x 4 x 3 x 2 x 1 = 120

2. How many 6 digit telephone numbers can be formed if each number starts with 35 and no digit appears more than once?A. 720 B. 360 C. 1420 D. 1680

Explanation : The first two places can only be filled by 3 and 5 respectively and there is only 1 way of doing this

Given that no digit appears more than once. Hence we have 8 digits remaining(0,1,2,4,6,7,8,9)So, the next 4 places can be filled with the remaining 8 digits in 8P4 ways

Total number of ways = 8P4 = 8 x 7 x 6 x 5 = 1680

3. An event manager has ten patterns of chairs and eight patterns of tables. In how many ways can he make a pair of table and chair?A. 100 B. 80 C. 110 D. 64

Explanation : He has has 10 patterns of chairs and 8 patterns of tables

Hence, A chair can be arranged in 10 ways and A table can be arranged in 8 waysHence one chair and one table can be arranged in 10 x 8 ways = 80 ways

4. In how many different ways can the letters of the word 'CORPORATION' be arranged so that the vowels always come together?A. 47200 B. 48000 C. 42000 D. 50400Explanation : The word 'CORPORATION' has 11 letters. It has the vowels 'O','O','A','I','O' in it and these 5 vowels should always come together. Hence these 5 vowels can be grouped and considered as a single letter. that is, CRPRTN(OOAIO).

Hence we can assume total letters as 7. But in these 7 letters, 'R' occurs 2 times andrest of the letters are different.Number of ways to arrange these letters = 7!/ 2!=7×6×5×4×3×2×1/ 2×1=2520

In the 5 vowels (OOAIO), 'O' occurs 3 and rest of the vowels are different.

Number of ways to arrange these vowels among themselves = 5!/ 3!= 5×4×3×2×1/ 3×2×1=20Hence, required number of ways = 2520 x 20 = 50400

5. How many 3 digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9 which are divisible by 5 and none of the digits is repeated?A. 20 B. 16 C. 8 D. 24

Explanation : A number is divisible by 5 if the its last digit is a 0 or 5We need to find out how many 3 digit numbers can be formed from the 6 digits (2,3,5,6,7,9)which are divisible by 5.Required Number of three digit numbers = 4 x 5 x 1 = 20

6. How many numbers are there between 100 and 1000 such that at least one of their digits is 6?(a) 648 (b) 258 (c) 654 (d) 252

7. How many numbers not exceeding 10000 can be made using the digits 2,4,5,6,8 if repetition of digits is allowed?A. 9999 B. 820 C. 780 D. 740

Explanation : Given that the numbers should not exceed 10000

Hence numbers can be 1 digit numbers or 2 digit numbers or 3 digit numbers or 4 digit numbers

Given that repetition of the digits is allowed.

A. Count of 1 digit numbers that can be formed using the 5 digits (2,4,5,6,8) (repetition allowed)

The unit digit can be filled by any of the 5 digits (2,4,5,6,8)total count of numbers not exceeding 10000 that can be made using the digits 2,4,5,6,8 (with repetition of digits) = 5 + 52 + 53 + 54 =780

8. 25 buses are running between two places P and Q. In how many ways can a person go from P to Q and return by a different bus?A. 50 B. 600 C. 576 D. 625

Explanation : He can go in any bus out of the 25 buses. Hence He can go in 25 ways.

Since he can not come back in the same bus that he used for travelling,He can return in 24 ways. Total number of ways = 25 x 24 = 600

9. In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?A. 159 B. 209 C. 201 D. 212

Total number of ways = (6C4) + (6C3 x 4C1) + (6C2 x 4C2) + (6C1 x 4C3) = (6C2) + (6C3 x 4C1) + (6C2 x 4C2) + (6C1 x 4C1) = 15 + 80 + 90 + 24 = 209

10. There are 6 periods in each working day of a school. In how many ways can one organize 5 subjects such that each subject is allowed at least one period?A. 3200 B. 3600 C. 2400 D. None of these

Explanation : We have 6 periods and need to organize 5 subjects such that each subject is allowed at least one period.we have 5 sub and 6 periods so their arrangement is 6P5 and now we have 1 period which we can fill with any of the 5 subjects so 5C1 6P5*5C1=3600

11. What is the sum of all 4 digit numbers formed using the digits 2, 3,4 and 5 without repetition?A. 93324 B. 92314 C. 93024 D. 91242

Explanation : If all the possible n digit numbers using the n distinct digits are formed, the sum of all the numbers so formed is equal to (n-1)! × (Sum of the n digits) × (111 ... n times)Hence the sum of all 4 digit numbers formed using the digits 2, 3,4 and 5 without repetition = (4-1)! (2 + 3 + 4 + 5)(1111) = 3! × 14 × 1111 = 6 × 14 × 1111 = 93324

Alter Method:Total no. formed = 4!=4*3*2*1=24Starting with each number is 6 times (24/4)(No of ways/ total number’s)Sum of digits starting with 1,2,5,6 = 6*1+6*2+6*5+6*6=6*(1+2+5+6)=84This sum will be same for hundred, tens & unit places

Sum of all 4 digit numbers formed using digits 1,2,5,6= 1000*84+100*84+10*84+84 =84*(1000+100+10+1)=84*1111 =93324

12. In a birthday party, every person shakes hand with every other person. If there was a total of 28 handshakes in the party, how many persons were present in the party?A. 9 B. 8 C. 7 D. 6

Assume that in a party every person shakes hand with every other person. Let n = the total number of persons present in the party

h = total number of handshakesThen, h=n(n−1)/2Here h = 28, h=n(n−1)/ 2⇒28=n(n−1)/ 2 => n(n-1) = 28 × 2 =>n(n-1)= 56 => n = 8

13. In how many different ways can 5 girls and 5 boys form a circle such that the boys and the girls alternate?A. 2880 B. 1400 C. 1200 D. 3212

Explanation : In a circle, 5 boys can be arranged in 4! ways

Given that the boys and the girls alternate. Hence there are 5 places for girls which can be arranged in 5! ways

Total number of ways = 4! x 5! = 24 x 120 = 2880

14. Find out the number of ways in which 6 rings of different types can be worn in 3 fingers?A. 120 B. 216 C. 125 D. 729

Explanation : The first ring can be worn in any of the 3 fingers=> There are 3 ways of wearing the first ringSimilarly each of the remaining 5 rings also can be worn in 3 waysHence total number of ways =3×3×3×3×3×3=36=729

15. If there are 9 horizontal lines and 9 vertical lines in a chess board, how many rectangles can be formed in the chess board?A. 920 B. 1024 C. 64 D. 1296

The number of rectangles that can be formed by using m horizontal lines and n vertical lines aremC2 × nC2

Here m = 9, n = 9Hence, The number of rectangles that can be formed = mC2 × nC2

= 9C2 × 9C2 = (9C2)2

=(9×8/ 2×1)2=362=1296

16. How many five digit positive integers that are divisible by 3 can be formed using the digits 0, 1, 2, 3, 4 and 5, without any of the digits getting repeating(a) 15 (b) 96 (c) 216 (d) 120

17. If the letters of the word CHASM are rearranged to form 5 letter words such that none of the word repeat and the results arranged in ascending order as in a dictionary what is the rank of the word CHASM?

A. 24 B. 31 C. 32 D. 30

Explanation : The 5 letter word can be rearranged in 5! Ways = 120 without any of the letters repeating.The first 24 of these words will start with A.

Then the 25th word will start will CA _ _ _. The remaining 3 letters can be rearranged in 3! Ways = 6. i.e. 6 words exist that start with CA.

The next word starts with CH and then A, i.e., CHA _ _. The first of the words will be CHAMS. The next word will be CHASM.

Therefore, the rank of CHASM will be 24 + 6 + 2 = 32.

18. How many four letter distinct initials can be formed using the alphabets of English language such that the last of the four words is always a consonant?

A. (26^3)*(21) B. 26*25*24*21 C. 25*24*23*21 D. None of these.

Explanation : The last of the four letter words should be a consonant. Therefore, there are 21 options.

The first three letters can be either consonants or vowels. So, each of them have 26 options. Note that the question asks you to find out the number of distinct initials and not initials where the letters are distinct.

Hence answer = 26*26*26*21 = 263 * 21

19. There are 12 yes or no questions. How many ways can these be answered? A. 1024 B. 2048 C. 4096 D. 144

Explanation : Each of the questions can be answered in 2 ways (yes or no)Therefore, no. of ways of answering 12 questions = 212 = 4096 ways.

20. A person writes letters to six friends and addresses the corresponding envelopes. In how many ways can the letters be placed in the envelopes so that at least two of them are in wrong envelopes?

(a) 119 (b) 120 (c) 720 (d) 719

21. There are 12 points out of which 7 points are marked on the straight line. How many triangles is it possible to form?

(a) 185 (b) 175 (c) 7! × 5! (d) 140

22. A seven-digit number comprises of only 2's and 3's. How many of these are multiples of 12?

(a) 1 (b) 21 (c) 11 (d) 47

Explanation :

Number should be a multiple of 3 and 4. So, the sum of the digits should be a multiple of 3. 

We can either have all seven digits as 3, or have three 2's and four 3's, or six 2's and a 3 (The number of 2's should be a multiple of 3).

For the number to be a multiple of 4, the last 2 digits should be 32. Now, let us combine these two.

(i)- All seven 3's - No possibility

(ii)- Three 2's and four 3's - The first 5 digits should have two 2's and three 3's in some order. No of possibilities =5!/3!*2!=10

(iii)- Six 2's and one 3 - The first 5 digits should all be 2's. So, there is - only one number - 2222232

So we get the total of seven-digit number comprises of only 2's and 3's which are multiples of 12 are: = (i) + (ii) + (iii) = 0+10+1=11

23. How many 5-digit positive integers exist the sum of whose digits are odd?

(a) 36000 (b) 38000 (c) 45000 (d) 90000

Explanation:

There are 9×104=90000 5-digit positive integers.

Out of these 90000 positive integers, the sum of the digits of half of the numbers will add up to an odd number and the remaining half will add up to an even number.

Hence there are 90000/2

= 45000 5-digit positive integers whose sum add up to an odd number.

24. A 6x6 grid is cut from an 8x8 chessboard. In how many ways can we put two identical coins, one on the black square and one on a white square on the grid, such that they are not placed in the same row or in the same column?

(a) 216 (b) 324 (c) 144 (d) 108

Explanation : In a 6x6 grid of a chessboard, each row and each column contains 3 white and 3 black squares placed alternatively. There are a total of 18 black and 18 white squares. For every black square chosen to put one coin, we cannot choose any white square present in its row or column. There are 3 white squares in its row and 3 white square in its column for every black square.

Hence for every black square chosen, we can choose (18−6)=12 white squares.

Total number of possibilities where a black square and a white square can be chosen so that they do not fall in the same row or in the same column: =18×12=216

So there are 216 ways of placing the coins that are identical.

25. In how many ways can the letters of the word EDUCATION be rearranged so that the relative position of the vowels and consonants remain the same as in the word EDUCATION?

(a)

9 !4 !×5 ! (b)

9 !4 ! (c)4! ×5! (d) None of these

Explanation : The word EDUCATION is a 9 letter word, with none of the letters repeating.The vowels occupy 3rd,5th,7th and 8th position in the word and the remaining 5 positions are occupied by consonantsAs the relative position of the vowels and consonants in any arrangement should remain the same as in the word EDUCATION, the vowels can occupy only the afore mentioned 4 places and the consonants can occupy1st,2nd,4th,6th and 9th positions.The 4 vowels can be arranged in the 3rd,5th,7th and 8th position in 4! Ways.Similarly, the 5 consonants can be arranged in 1st,2nd,4th,6th and 9th position in 5! Ways.Hence, the total number of ways = 4! × 5!

ASSIGNMENT

1. The sum of all the numbers which can be formed by using the digits 1, 2, 3, 4 all at a time (no digit is being repeated) is

(a) 10000 (b) 60000 (c) 66000 (d) 66660

2. In how many ways can the letters of the word ‘MACHINE’ be arranged so that the vowels may occupy only odd positions?

(a) 4 x 7 (b) 576 (c) 288 (d) none

3. The number of positive integers greater than 6000 and less than 7000 which are divisible by 5, with no digit repeated is

(a) 28 (b) 56 (c) 112 (d) 84

4. How many words can be formed with the letters of the word “PATALIPUTRA” without changing the relative order of the vowels and consonants?

(a) 3600 (b) 3900 (c) 6300 (d) 4600

5. In a symposium, four speakers – an economist, a politician, a union leader and a social worker, are to speak one after another. How many schedules showing the order in which these persons speak will have the economist speaking always after the politician?

(a) 6 (b) 12 (c) 24 (d) None of these

6. How many words can be formed using all the letters of the word ‘LAUGHTER’ so that the vowels are never together?

(a) 14400 (b) 3600 (c) 40320 (d) 36000

7. If there are twelve persons in a party and if each two of them shake hands with each other, how many hand shakes happen in the party?

(a) 66 (b) 55 (c) 24 (d) 48

8. How many different Barbie’s can I make if I can choose from 4 different skin colours, 3 different eye colours, 4 different hairstyles and 6 different hair colours?

(a) 17 (b) 288 (c) 2488320 (d) None of these

9. A committee of seven is to be formed from 9 boys and 5 girls. In how many ways can this be done, when a committee contains at least three girls?

(a) 1726 (b) 1716 (c) 1617 (d) 2617

10. How many 4 letter words with or without meaning can be formed out of the letters of the word ‘LOGARITHMS’, if repetition of letters is not allowed?

(a) 40 (b) 400 (c) 5040 (d) 2520

11. How many different numbers greater than three lakhs can be formed out of the digits 111223?

(a) 60 (b) 10 (c) 120 (d) 720

12. In how many ways can 21 books on English and 19 books on Hindi be placed in a row on a shelf so that two books on Hindi may not be together?

(a) 3990 (b) 1540 (c) 1995 (d) 3672

13. In how many ways 4 boys and 4 girls can be seated in a row so that boys and girls are alternate?

(a) 57 60 (b) 5040 (c) 40320 (d) 1152

14. Find the number of numbers between 300 and 3000 that can be formed with the digits 0,1,2,3,4 and 5, no digits being repeated in any number

(a) 90 (b) 120 (c) 160 (d) 180

15. Five Matches are to be played in a chess tournament. In how many ways can their results be decided?

(a) 243 (b) 343 (c) 128 (d) none of these