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1.

^4Pr = 4*^5Pr-1. Find r.(a) 1(b) 2(c) 3(d) 4The question was posed to me during an online interview.My question is based upon Permutations-2 in division Permutations and Combinations of Mathematics – Class 11

Answer»

The correct ANSWER is (a) 1

The best explanation: ^4PR = 4*^5Pr-1

=> \(\frac{4!}{(4-r)!} = 4*\frac{5!}{(5-r+1)!}\)

=> \(\frac{(6-r)!}{(4-r)!} = 4*\frac{5!}{4!}\)

=> (6-r) (5-r) = 4*5

=> r=1.

2.

Find the number of 5 letter words which can be formed from word IMAGE without repetition using permutations.(a) 20(b) 60(c) 120(d) 240The question was asked in class test.My doubt stems from Permutations-1 topic in section Permutations and Combinations of Mathematics – Class 11

Answer»

The correct ANSWER is (c) 120

Easy explanation: IMAGE is a 5 letters word. We have to arrange all 5 letters of the word IMAGE without REPETITION. So, total PERMUTATIONS are ^nPr = ^5P5 = 5! = 5.4.3.2.1 = 120.

3.

The number of permutations of n different objects taken r at a time, where repetition is allowed is _______________(a) n!(b) r!(c) ^nPr(d) n^rThe question was asked in semester exam.This interesting question is from Permutations-1 in portion Permutations and Combinations of Mathematics – Class 11

Answer»

Right answer is (d) N^R

The best explanation: The NUMBER of permutations of n different objects TAKEN r at a time, where repetition is allowed is n*n*n*n*n……………. r TIMES = n^r.

4.

A child has 2 pencil and 3 erasers. In how many ways he can take a pencil and an eraser?(a) 5(b) 6(c) 8(d) 9This question was addressed to me by my school principal while I was bunking the class.This is a very interesting question from Fundamental Principle of Counting topic in section Permutations and Combinations of Mathematics – Class 11

Answer»

The correct choice is (b) 6

To explain I would say: By the fundamental principle of COUNTING, if an event can occur in ‘m’ different ways, following which another event can occur in ‘n’ different ways, then the total numbers of occurrence of the EVENTS in the given order is m*n.

So, if pencil can be taken in 2 ways and ERASER can be taken in 3 ways then number of ways in which he can TAKE a pencil and eraser are 2*3 = 6.

5.

In a family, 5 males and 3 females are there. In how many ways we can select a group of 2 males and 2 females from the family?(a) 3(b) 10(c) 30(d) 40The question was asked by my college director while I was bunking the class.My query is from Combinations in chapter Permutations and Combinations of Mathematics – Class 11

Answer»

Right choice is (c) 30

Best explanation: Out of 5 males, 2 males can be selected in ^5C2 ways.

^5C2 = \(\frac{5!}{(5-2)! 2!} = \frac{5*4*3!}{(3)! 2!} = \frac{20}{2}\) = 10 ways.

Out of 3 females, 2 females can be selected in ^3C2 ways .

^3C2 = \(\frac{3!}{(3-2)! 2!} = \frac{3*2!}{(1)! 2!} = \frac{3}{1}\) = 3 ways.

So, by the FUNDAMENTAL principle of counting we can SELECT a GROUP of 2 males and 2 females from the family in 10*3 = 30 ways.

6.

Out of a group of 5 persons, find the number of ways of selecting 3 persons.(a) 1(b) 5(c) 10(d) 15I got this question in unit test.This interesting question is from Combinations in division Permutations and Combinations of Mathematics – Class 11

Answer»

Correct ANSWER is (c) 10

Explanation: Out of a group of 5 persons, we have to select 3 persons. This can be done in ^5C3 ways.

^nCr = \(\frac{n!}{(n-R)! r!}\)

^5C3 = \(\frac{5!}{(5-3)! 3!} = \frac{5*4*3!}{(2)! 3!} = \frac{20}{2}\)

i.e. 10 ways.

7.

^nPr = ^nCr * ______________(a) r!(b) 1/r!(c) n!(d) 1/n!I had been asked this question by my school principal while I was bunking the class.My doubt stems from Combinations in portion Permutations and Combinations of Mathematics – Class 11

Answer»

The CORRECT choice is (a) r!

The best I can explain: We know, ^NPR = \(\FRAC{n!}{(n-r)!}\) and ^nCr = \(\frac{n!}{(n-r)! r!}\).

=> ^nPr / ^nCr = \(\frac{\frac{n!}{(n-r)!}}{\frac{n!}{(n-r)! r!}}\) = r!

=> ^nPr = ^nCr * r!

8.

Order matters in combination.(a) True(b) FalseThe question was asked in unit test.The above asked question is from Combinations topic in division Permutations and Combinations of Mathematics – Class 11

Answer»

Correct ANSWER is (b) False

To elaborate: Combination MEANS SELECTION and selection does not require ordering. So, ORDER does not MATTER in combination.

9.

If there are 4 paths to travel from Delhi to Kanpur, then in how many ways a person can travel from Delhi to Kanpur and came back to Delhi via same path?(a) 4(b) 8(c) 12(d) 16The question was posed to me in quiz.My question is taken from Fundamental Principle of Counting in portion Permutations and Combinations of Mathematics – Class 11

Answer»

Correct option is (a) 4

To explain: If there are 4 paths to TRAVEL from Delhi to Kanpur and person take one path then in COMING BACK journey only one option is there. So, by the fundamental principle of counting, total NUMBER of ways a person can travel from Delhi to Kanpur and came back to Delhi via same path are 4*1 = 4.

10.

^nPn = ________________(a) n!(b) 1(c) \(\frac{1}{(n)!}\)(d) (n-1)!The question was posed to me in examination.My doubt is from Permutations-1 in portion Permutations and Combinations of Mathematics – Class 11

Answer»

The CORRECT ANSWER is (a) n!

The EXPLANATION: We know, ^NPR = \(\frac{n!}{(n-r)!}\).

^nPn = \(\frac{n!}{(n-n)!} = \frac{n!}{(0)!}\) = n!

11.

^nCn = ________________(a) n!(b) 1(c) \(\frac{1}{(n)!}\)(d) (n-1)!The question was asked by my school teacher while I was bunking the class.This interesting question is from Combinations in section Permutations and Combinations of Mathematics – Class 11

Answer»

Correct answer is (b) 1

The best I can EXPLAIN: We know, ^NCR = \(\frac{n!}{(n-R)! r!}\).

^nCn = \(\frac{n!}{(n-n)! n!} = \frac{n!}{(0)! n!}\) = 1/1 = 1.

12.

How many 5-digit numbers are possible without repetition of digits?(a) 27216(b) 50400(c) 100000(d) 90000I had been asked this question in an interview for internship.I'd like to ask this question from Fundamental Principle of Counting in chapter Permutations and Combinations of Mathematics – Class 11

Answer»

The correct choice is (a) 27216

For EXPLANATION I would say: TOTAL DIGITS are 0-9 i.e.10. In a 5-digit number 0 cannot be at first place.

So, by the fundamental principle of counting, total numbers possible are 9*9*8*7*6 = 27216.

13.

How many 4 digits even numbers are possible from digits 1 to 9 if repetition is not allowed?(a) 6561(b) 2016(c) 1344(d) 2916This question was posed to me in an online quiz.My doubt stems from Fundamental Principle of Counting in chapter Permutations and Combinations of Mathematics – Class 11

Answer»

Correct answer is (c) 1344

Explanation: Total digits are 1-9 i.e.9. Even NUMBER means 2,4,6,8 at unit’s place.

So, 4 possible digits for unit place, after a digit is placed at unit place 8 digits are remaining.

8 ways to fill 1st POSITION, 7 ways to fill 2nd position, 6 ways to fill th3 3rd position.

Hence, by the fundamental principle of COUNTING, total possible numbers are 8*7*6*4 = 1344.

14.

^nC0 = ________________(a) n!(b) 1(c) \(\frac{1}{(n)!}\)(d) (n-1)!I got this question in a national level competition.I want to ask this question from Combinations in section Permutations and Combinations of Mathematics – Class 11

Answer»

Correct choice is (b) 1

The explanation: We KNOW, ^nCr = \(\FRAC{n!}{(n-R)! r!}\).

^nC0 = \(\frac{n!}{(n-0)! 0!} = \frac{n!}{n!(1)}\) = 1.

15.

Find the number of 4 letter words which can be formed from word IMAGE if repetition is allowed.(a) 120(b) 125(c) 625(d) 3125I got this question by my school teacher while I was bunking the class.My question is based upon Permutations-2 topic in portion Permutations and Combinations of Mathematics – Class 11

Answer»

The correct answer is (c) 625

The BEST EXPLANATION: We have to ARRANGE 4 letters of the 5 letters word IMAGE without repetition. So, TOTAL permutations are n^r = 5^4 = 625.

16.

How many 4 digits even numbers are possible from digits 1 to 9 if repetition is allowed?(a) 6561(b) 2016(c) 1344(d) 2916I have been asked this question in an interview for internship.The query is from Fundamental Principle of Counting topic in division Permutations and Combinations of Mathematics – Class 11

Answer»

The correct CHOICE is (d) 2916

The best explanation: Total digits are 1-9 i.e.9. EVEN number means 2,4,6,8 at unit’s PLACE.

So, 4 possible digits for unit place. Rest all positions are FILLED in 9 ways.

Hence, by the fundamental principle of COUNTING, total possible numbers are 9*9*9*4 = 2916.

17.

Permutation is also known as selection.(a) True(b) FalseThis question was addressed to me during an online exam.The doubt is from Permutations-1 in chapter Permutations and Combinations of Mathematics – Class 11

Answer»

The correct option is (b) False

The BEST I can explain: PERMUTATION is known as arrangement. SELECTION is another name for combinations.

It INVOLVES arrangement of letters, numbers, persons ETC.

18.

If ^6C2 = ^6CX then find possible values of x.(a) 2(b) 4(c) 2 and 4(d) 3The question was asked in final exam.My doubt stems from Combinations in section Permutations and Combinations of Mathematics – Class 11

Answer» CORRECT choice is (C) 2 and 4

Easiest EXPLANATION: We KNOW, ^nCp = ^nCq

=>either p = q or p + q = n.

=> either x=2 or x+2=6

=> either x=2 or x=4.
19.

Is ^nCr = ^nCn-r true?(a) True(b) FalseI had been asked this question at a job interview.The doubt is from Combinations in chapter Permutations and Combinations of Mathematics – Class 11

Answer»

Right CHOICE is (a) True

The explanation: We know, ^NCR = \(\frac{N!}{(n-r)! r!}\).

REPLACING r by n-r, we get ^nCn-r = \(\frac{n!}{(n-(n-r))!(n-r)!} = \frac{n!}{r! (n-r)!}\) = ^nCr

=> ^nCr = ^nCn-r

20.

If ^nP3 = 4*^nP2. Find n.(a) 3(b) 2(c) 6(d) 5The question was posed to me by my college director while I was bunking the class.I want to ask this question from Permutations-2 topic in section Permutations and Combinations of Mathematics – Class 11

Answer»

Right option is (c) 6

Explanation: ^nP3 = 4*^nP2

\(\FRAC{n!}{(n-3)!} = 4 * \frac{n!}{(n-2)!}\)

=> n-2 = 4

=> n=6.

21.

^nP0 = ________________(a) n!(b) 1(c) \(\frac{1}{(n)!}\)(d) (n-1)!I got this question in an online interview.My doubt is from Permutations-1 topic in division Permutations and Combinations of Mathematics – Class 11

Answer»

The CORRECT answer is (B) 1

The explanation: We know, ^nPr = \(\FRAC{N!}{(n-r)!}\).

^nP0 = \(\frac{n!}{(n-0)!} = \frac{n!}{(n)!}\) = 1.

22.

\(\frac{7!}{5!}\) = ____________________(a) 7(b) 42(c) 230(d) 30The question was posed to me in final exam.The query is from Permutations-1 in portion Permutations and Combinations of Mathematics – Class 11

Answer»

The CORRECT answer is (b) 42

To explain I would say: We know, N! = n.(n-1).(n-2).(n-3)…… = n(n-1)!

\(\FRAC{7!}{5!} = \frac{7.6!}{5!} = \frac{7.6.5!}{5!}\) = 7.6 = 42.

23.

6! = _____________(a) 24(b) 120(c) 720(d) 8This question was addressed to me during an interview.My question is from Permutations-1 in portion Permutations and Combinations of Mathematics – Class 11

Answer» RIGHT option is (c) 720

Easy explanation: We KNOW, N! = n.(n-1).(n-2).(n-3)…..

6! = 6.5.4.3.2.1 = 720.
24.

^nPr = ________________(a) n!(b) \(\frac{n!}{r!}\)(c) \(\frac{n!}{(n-r)!}\)(d) \(\frac{n!}{(n-r)! r!}\)The question was posed to me by my college director while I was bunking the class.I would like to ask this question from Permutations-1 topic in chapter Permutations and Combinations of Mathematics – Class 11

Answer» CORRECT ANSWER is (C) \(\frac{n!}{(n-r)!}\)

The best I can EXPLAIN: Permutation is known as arrangement. ^nPr MEANS arranging r objects out of n.

^nPr = \(\frac{n!}{(n-r)!}\).
25.

Find the number of different 8-letter arrangements that can be made from the letters of the word EDUCATION so that all vowels occur together.(a) 40320(b) 37440(c) 1440(d) 2880I got this question in a job interview.My question is from Permutations-2 topic in chapter Permutations and Combinations of Mathematics – Class 11

Answer»

The correct answer is (d) 2880

Explanation: There are 5 vowels in WORD EDUCATION. 5 vowels can be arranged in ^5P5 i.e. 5! ways. When all vowels are together, 5 vowels together FORM one LETTER and remaining 3 letters i.e. together 4 letters can be arranged in ^4P4 i.e. 4! ways. Total possible arrangements are 5! * 4! = 120*24 = 2880.

26.

If a code involves one letter of English alphabet at first place and a number at second place. Numbers can be used from 1 to 9. How many codes are possible using the letters of alphabet and numbers?(a) 234(b) 260(c) 314(d) 324I have been asked this question during an internship interview.My question is from Fundamental Principle of Counting topic in division Permutations and Combinations of Mathematics – Class 11

Answer»

The correct answer is (a) 234

Easiest explanation: Since there are 26 LETTERS in English ALPHABET and 9 possible DIGITS from 1 to 9.

So, by the fundamental principle of counting, total possible codes are 26*9 = 234.

27.

Find the number of 5 letter words which can be formed from word PULSE if repetition is allowed.(a) 25(b) 120(c) 125(d) 3125I have been asked this question in my homework.The origin of the question is Fundamental Principle of Counting topic in division Permutations and Combinations of Mathematics – Class 11

Answer»

Right OPTION is (d) 3125

The BEST explanation: PULSE is a 5 LETTERS word. By the fundamental principle of counting, total NUMBER of possible words with repetition allowed are 5*5*5*5*5 = 3125.

28.

If ^14Cr = 14 and ^15Cr = 15. Find the value of ^14Cr-1.(a) 1(b) 14(c) 15(d) 3The question was posed to me by my college director while I was bunking the class.I would like to ask this question from Combinations topic in chapter Permutations and Combinations of Mathematics – Class 11

Answer»

The correct ANSWER is (a) 1

For explanation I WOULD SAY: We know, ^nCr + ^nCr-1 = ^n+1Cr

Substituting n=14 and r=4, we GET ^14Cr + ^14Cr-1 = ^15Cr

=>^14Cr-1 = ^15Cr – ^14Cr = 15-14 = 1.

29.

If ^nC2 = ^nC3 then find n.(a) 2(b) 3(c) 5(d) 6I have been asked this question in an online quiz.I want to ask this question from Combinations topic in section Permutations and Combinations of Mathematics – Class 11

Answer»

Correct CHOICE is (c) 5

To elaborate: We know, ^NCP = ^nCq

=>either p = q or p + q = n.

Here p=2 and q=3 so, p ≠ q.

Hence p + q = n => n = p + q = 2 + 3 = 5.

30.

How many 3-digit numbers are possible using permutations without repetition of digits if digits are 1-9?(a) 504(b) 729(c) 1000(d) 720I got this question in an interview for job.My query is from Permutations-2 in portion Permutations and Combinations of Mathematics – Class 11

Answer»

Right choice is (a) 504

Easiest explanation: 1-9 digits which MEANS 9 digits are possible. We have to ARRANGE 3 digits at a time out of 9 digits WITHOUT repetition. So, TOTAL permutations are ^nPr = ^9P3 = \(\frac{9!}{(9-3)!} = \frac{9!}{6!}\) = 9.8.7 = 504.

31.

A passcode is made of 5 digits. How many maximum numbers of ways incorrect passcode is entered?(a) 27215(b) 50399(c) 99999(d) 89999The question was posed to me in class test.This question is from Fundamental Principle of Counting in portion Permutations and Combinations of Mathematics – Class 11

Answer»

Correct answer is (C) 99999

For EXPLANATION I would say: TOTAL digits are 0-9 i.e.10. In passcode, 0 may also OCCUR at first place.

All places are filled in 10 ways.

So, by the FUNDAMENTAL principle of counting, total numbers possible are 10*10*10*10*10=100000.

Maximum number of incorrect pass code entered = 100000-1 = 99999.

32.

Determine n if ^2nC3: ^nC3 = 9:1.(a) 7(b) 14(c) 28(d) 32I have been asked this question in a job interview.I'm obligated to ask this question of Combinations topic in portion Permutations and Combinations of Mathematics – Class 11

Answer»

Correct answer is (b) 14

The BEST I can explain: We know, ^nCr = \(\frac{N!}{(n-r)! r!}\).

^2nC3 / ^nC3 = \(\frac{\frac{n!}{3! (2N-3)!}}{\frac{n!}{3! (n-3)!}}\)

= \(\frac{2n! (n-3)!}{n! (2n-3)!}\)

9/1 = \(\frac{2n(2n-1)(2n-2)}{n(n-1)(n-2)}\)

9 = \(\frac{2(2n-1)2}{(n-2)}\)

9n-18 = 8n-4

=>n=14.

33.

^nCr = ________________(a) n!(b) \(\frac{n!}{r!}\)(c) \(\frac{n!}{(n-r)!}\)(d) \(\frac{n!}{(n-r)! r!}\)This question was addressed to me during an online exam.This intriguing question comes from Combinations in chapter Permutations and Combinations of Mathematics – Class 11

Answer»

Correct answer is (d) \(\FRAC{n!}{(n-R)! r!}\)

The best I can explain: Permutation is known as SELECTION. ^NCR means arranging r OBJECTS out of n.

^nCr = \(\frac{n!}{(n-r)! r!}\).

34.

Find the number of 5 letter words that can be formed from word IMAGE using permutations if repetition is allowed.(a) 25(b) 120(c) 125(d) 3125This question was posed to me in quiz.The origin of the question is Permutations-1 topic in portion Permutations and Combinations of Mathematics – Class 11

Answer»

Right option is (d) 3125

The best I can EXPLAIN: IMAGE is a 5 letters WORD. We have to arrange all 5 letters of the word IMAGE with REPETITION allowed. So, total permutations are n^r = 5^5 = 3125.

35.

Find the number of permutations of word DEPENDENT.(a) 132400(b) 1512500(c) 1663200(d) 1723400I have been asked this question in quiz.The question is from Permutations-1 in division Permutations and Combinations of Mathematics – Class 11

Answer»

Correct CHOICE is (c) 1663200

To explain: There are total 9 letters out of which 1T, 2N, 2D, 3E, 1P.

Total number of permutations are \(\FRAC{9!}{3!2!2!} = \frac{9.8.7.6.5.4.3.2.1}{6*2*2} = \frac{362880}{24}\) = 15120.

36.

Find the number of 4 letter words which can be formed from word PULSE if repetition is allowed.(a) 120(b) 125(c) 625(d) 3125I got this question during a job interview.The query is from Fundamental Principle of Counting topic in section Permutations and Combinations of Mathematics – Class 11

Answer»

Correct option is (C) 625

To elaborate: PULSE is a 5 letters word. By the fundamental PRINCIPLE of counting, total number of possible WORDS with repetition ALLOWED are 5*5*5*5 = 625.

37.

If there are 4 paths to travel from Delhi to Kanpur, then in how many ways a person can travel from Delhi to Kanpur and came back to Delhi via different path?(a) 4(b) 8(c) 12(d) 16This question was posed to me during an interview for a job.My enquiry is from Fundamental Principle of Counting in section Permutations and Combinations of Mathematics – Class 11

Answer»

The correct choice is (c) 12

The best explanation: If there are 4 paths to TRAVEL from DELHI to Kanpur and person take one path then in coming back journey only 3 paths are remaining. So, by the fundamental principle of counting, total number of ways a person can travel from Delhi to Kanpur and came back to Delhi via different path are 4 * 3 = 12.

38.

How many 3-digit numbers are possible using permutations with repetition allowed if digits are 1-9?(a) 504(b) 729(c) 1000(d) 720The question was asked in quiz.This intriguing question originated from Permutations-2 in portion Permutations and Combinations of Mathematics – Class 11

Answer»

The correct CHOICE is (b) 729

To explain I WOULD SAY: 1-9 digits which means 9 digits are possible. We have to arrange 3 digits at a time out of 9 digits with repetition allowed. So, total PERMUTATIONS are n^r = 9^3 = 729.

39.

\(\frac{100}{10!} = \frac{1}{8!} + \frac{x}{9!}\). Find x.(a) 1(b) 2(c) 3(d) 4This question was addressed to me by my school teacher while I was bunking the class.The origin of the question is Permutations-1 topic in section Permutations and Combinations of Mathematics – Class 11

Answer»

Correct option is (a) 1

Easiest explanation: \(\frac{100}{10!} = \frac{1}{8!} + \frac{x}{9!}\).

We know, n! = n.(n-1). (n-2). (n-3) …………… = n(n-1)!

\(\frac{100}{10.9.8!} = \frac{1}{8!} + \frac{x}{9.8!}\)

=> \(\frac{100}{10.9} = \frac{1}{1} + \frac{x}{9}\)

=> \(\frac{10}{9} = 1 + \frac{x}{9}\)

=> \(\frac{x}{9} = \frac{10}{9} – 1 = \frac{1}{9}\)

=> x=1.

40.

Find the number of 5 letter words which can be formed from word PULSE without repetition.(a) 20(b) 60(c) 120(d) 240This question was addressed to me at a job interview.My query is from Fundamental Principle of Counting topic in chapter Permutations and Combinations of Mathematics – Class 11

Answer»

Correct ANSWER is (c) 120

Explanation: PULSE is a 5 letters word. By the FUNDAMENTAL principle of counting, TOTAL number of possible words without repetition are 5*4*3*2*1 = 120.

41.

How many 5-digit numbers are possible if repetition of digits is allowed?(a) 27216(b) 50400(c) 100000(d) 90000I have been asked this question during an interview for a job.The question is from Fundamental Principle of Counting topic in division Permutations and Combinations of Mathematics – Class 11

Answer»

The correct choice is (d) 90000

For explanation: Total DIGITS are 0-9 i.e.10. In a 5-digit number 0 cannot be at FIRST place.

So, by the fundamental PRINCIPLE of COUNTING, total numbers POSSIBLE are 9*10*10*10*10=90000.

42.

If there are 4 paths to travel from Delhi to Kanpur, then in how many ways a person can travel from Delhi to Kanpur and came back to Delhi?(a) 4(b) 8(c) 12(d) 16I had been asked this question by my school principal while I was bunking the class.The origin of the question is Fundamental Principle of Counting in section Permutations and Combinations of Mathematics – Class 11

Answer»

The correct option is (d) 16

The explanation is: If there are 4 paths to TRAVEL from DELHI to Kanpur then NUMBER of paths to travel from Kanpur to Delhi are 4. So, by the fundamental PRINCIPLE of counting, TOTAL number of ways to go and come back are 4*4 = 16.

43.

Find the number of words which can be made using all the letters of the word IMAGE. If these words are written as in a dictionary, what will be the rank of MAGIE?(a) 97(b) 98(c) 99(d) 100I had been asked this question in quiz.I need to ask this question from Permutations-2 in section Permutations and Combinations of Mathematics – Class 11

Answer»

Right answer is (c) 99

Easiest explanation: Words STARTING with letter A comes first in dictionary. Starting with A, number of words = 4! = 24. Starting with E, number of words = 4! = 24. Starting with I, number of words = 4! = 24. Starting with G, number of words = 4! = 24. SINCE our word also start with M so, we have to CONSIDER one more letter i.e. MA. Since our word also start with MA so, we have to consider one more letter i.e. MAE. Starting with MAE, number of words = 2! = 2. Since our word also start with MAG so, we have to consider one more letter i.e. MAGE. Starting with MAGE, only one letter i.e. MAGEI. After this, MAGIE comes. Total number of words before MAGIE = 24 + 24 + 24 + 24 + 2 = 98. So, rank of MAGIE is 99.

44.

In how many ways 2 red pens, 3 blue pens and 4 black pens can be arranged if same color pens are indistinguishable?(a) 362880(b) 1260(c) 24(d) 105680This question was posed to me by my school teacher while I was bunking the class.The question is from Permutations-2 in division Permutations and Combinations of Mathematics – Class 11

Answer»

Correct answer is (B) 1260

To explain: Total NUMBER of pens are 2+3+4 = 9 out of which 2 are of 1 type, 3 are of 2^nd type and 4 are of 3^rd type so, total number of arrangements = \(\frac{9!}{2!3!4!} = \frac{9*8*7*6*5}{2*6}\) = 1260.

45.

If an event can occur in ‘m’ different ways, following which another event can occur in ‘n’ different ways, then the total numbers of occurrence of the events in the given order is __________________(a) m + n(b) m – n(c) m*n(d) m/nThis question was addressed to me in an interview for internship.My question is based upon Fundamental Principle of Counting in chapter Permutations and Combinations of Mathematics – Class 11

Answer» CORRECT CHOICE is (C) m*n

Easy explanation: By the fundamental principle of counting, if an event can occur in ‘m’ DIFFERENT ways, following which another event can occur in ‘n’ different ways, then the total numbers of OCCURRENCE of the events in the given order is m*n.
46.

Find the number of 4 letter words which can be formed from word PULSE without repetition.(a) 20(b) 60(c) 120(d) 240I have been asked this question in an international level competition.My question is from Fundamental Principle of Counting in section Permutations and Combinations of Mathematics – Class 11

Answer»

The correct answer is (c) 120

Easy explanation: PULSE is a 5 LETTERS word. By the fundamental principle of counting, total NUMBER of possible words without REPETITION are 5*4*3*2 = 120.

47.

Find the number of 4 letter words which can be formed from word IMAGE using permutations without repetition.(a) 20(b) 60(c) 120(d) 240The question was posed to me in class test.My question is taken from Permutations-2 in portion Permutations and Combinations of Mathematics – Class 11

Answer»

Right choice is (c) 120

The BEST I can EXPLAIN: We have to arrange 4 letters of the 5 letters word IMAGE without repetition. So, total PERMUTATIONS are ^NPR = ^5P4 = 5! / (5-4)! = 5! = 5.4.3.2.1 = 120.

48.

Find the number of different 8-letter arrangements that can be made from the letters of the word EDUCATION so that all vowels do not occur together.(a) 40320(b) 37440(c) 1440(d) 2880I have been asked this question in examination.The origin of the question is Permutations-2 topic in division Permutations and Combinations of Mathematics – Class 11

Answer»

Correct CHOICE is (b) 37440

To explain: There are 5 VOWELS in word EDUCATION. 5 vowels can be arranged in ^5P5 i.e. 5! WAYS. When all vowels are TOGETHER, 5 vowels together form one letter and remaining 3 letters i.e. together 4 letters can be arranged in ^4P4 i.e. 4! ways. Total possible arrangements are 5! * 4! = 120*24 = 2880.

So, when all vowels do not occur together, total possible arrangements = 8! – 2880 = 40320 – 2880 = 37440.