Permutation And Combination — Part 1 MCQs with Explanations

Taking all person of same nationality as one person, then we will have only three people. These three person can be arranged themselves in 3! Ways. 8 Indians can be arranged themselves in 8! Way. 4 American can be arranged themselves in 4! Ways. 4 Englishman can be arranged themselves in 4! Ways. Hence, required number of ways = 3! 8! 4! 4! Ways.

APPLE = 5 letters. But two letters PP is of same kind. Thus, required permutations, $$\eqalign{ & = \frac{{5!}}{{2!}} \cr & = \frac{{120}}{2} \cr & = 60 \cr} $$

ALLAHABAD = 9 letters. Out of these 9 letters there is 4 A's and 2 L's are there. So, permutations = $$\frac{{9!}}{{4!.2!}}$$ = 7560 (a) There are 4 vowels and all are alike i.e. 4A's. _2 nd _4 th _6 th _8 th _ These even places can be occupied by 4 vowels. In $$\frac{{4!}}{{4!}}$$ = 1 Way. In other five places 5 other letter can be occupied of which two are alike i.e. 2L's. Number of ways = $$\frac{{5!}}{{2!}}$$ Ways. Hence, total number of ways in which vowels occupy the even places = $$\frac{{5!}}{{2!}}$$ × 1 = 60 ways. (b) Taking both L's together and treating them as one letter we have 8 letters out of which A repeats 4 times and others are distinct. These 8 letters can be arranged in $$\frac{{8!}}{{4!}}$$ = 1680 ways. Also two L can be arranged themselves in 2! ways. So, Total no. of ways in which L are together = 1680 × 2 = 3360 ways. Now, Total arrangement in which L never occur together, = Total arrangement - Total no. of ways in which L occur together. = 7560 - 3360 = 4200 ways

The question asks for the number of ways to arrange 10 examination papers so that the best and the worst papers never come together . This type of problem, where certain items "never come together", has a common strategy: Total Arrangements - Arrangements Where They ARE Together = Arrangements Where They Are NEVER Together Let's break it down step-by-step: Step 1: Calculate the Total Number of Ways to Arrange all 10 Papers. If we have 10 different papers, and we want to arrange them in a line, the number of ways is given by 10 factorial (10!) . 10! means 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1. So, Total arrangements = 10! Step 2: Calculate the Number of Ways where the Best and Worst Papers ARE Together. To make sure the best and worst papers are always together, we can think of them as a single "block" or "unit" . Imagine tying the best paper and the worst paper together. Now, instead of 10 individual papers, you have: (Paper 1), (Paper 2), ..., (Paper 8), and (Best & Worst Paper Block) . This means you now have 9 items to arrange (8 individual papers + 1 block). These 9 items can be arranged in 9 factorial (9!) ways. BUT, inside the "Best & Worst Paper Block", the best paper and the worst paper can swap their positions! They can be arranged in 2 factorial (2!) ways: (Best, Worst) or (Worst, Best) So, the total number of arrangements where the best and worst papers are together is: 9! × 2! Step 3: Calculate the Number of Ways where the Best and Worst Papers NEVER Come Together. Now we use our strategy: Never Together = Total Arrangements - Arrangements Where They ARE Together Never Together = 10! - (9! × 2!) Step 4: Simplify the Expression. We know that 10! can also be written as 10 × 9! And 2! is 2 × 1 = 2 . So, our equation becomes: Never Together = (10 × 9!) - (9! × 2) You can see that 9! is common in both parts. Let's factor it out: Never Together = 9! × (10 - 2) Never Together = 9! × 8 Or, simply, 8 × 9! This matches Option A . Alternative way No. of ways in which 10 paper can arranged is 10! Ways. When the best and the worst papers come together, regarding the two as one paper, we have only 9 papers. These 9 papers can be arranged in 9! Ways. And two papers can be arranged themselves in 2! Ways. No. of arrangement when best and worst paper do not come together, = 10! - 9! × 2! = 9!(10 - 2) = 8 × 9!

Let the Arrangement be, B G B G B G B 4 boys can be seated in 4! Ways Girl can be seated in 3! Ways Required number of ways, = 4! × 3! = 144

5 C 1 × 3 C 1 - 1 = 15 - 1 = 14

Number of ways $$\eqalign{ & { = ^{20}}{{\text{C}}_2} \cr & = \frac{{20!}}{{2! \times 18!}} \cr & = 20 \times \frac{{19}}{2} \cr & = 190 \cr} $$

There are 6 candidates and a voter has to vote for any two of them. So, the required number of ways is, $$\eqalign{ & { = ^6}{{\text{C}}_2} \cr & = \frac{{6!}}{{2! \times 4!}} \cr & = 15 \cr} $$

There 10 questions in part A out of which 8 question can be chosen as = 10 C 8 Similarly, 5 questions can be chosen from 10 questions of Part B as = 10 C 5 Hence, total number of ways, $$\eqalign{ & { = ^{10}}{{\text{C}}_8}{ \times ^{10}}{{\text{C}}_5} \cr & = \frac{{10!}}{{2! \times 8!}} \times \frac{{10!}}{{5! \times 5}} \cr & = \left\{ {10 \times \frac{9}{2}} \right\} \times \left\{ {\frac{{10 \times 9 \times 8 \times 7 \times 6}}{{5 \times 4 \times 3 \times 2 \times 1}}} \right\} \cr & = 11340 \cr} $$

A triangle needs 3 points. And polygon of 8 sides has 8 angular points. Hence, number of triangle formed, $$\eqalign{ & { = ^8}{{\text{C}}_3} \cr & = \frac{{8 \times 7 \times 6}}{{1 \times 2 \times 3}} \cr & = 56 \cr} $$

The number of triangle can be formed by 10 points = 10 C 3 Similarly, the number of triangle can be formed by 4 points when no one is collinear = 4 C 3 In the question, given 4 points are collinear, Thus, required number of triangle can be formed, = 10 C 3 - 4 C 3 = 120 - 4 = 116

Let n be the number of persons in the party Number of hands shake = 105 Total number of hands shake is given by n C 2 Now, According to the question, $$\eqalign{ & ^n{{\text{C}}_2} = 105 \cr & {\text{or, }}\frac{{n!}}{{2! \times \left( {n - 2} \right)!}} = 105 \cr & {\text{or, }}\frac{{n \times \left( {n - 1} \right)}}{2} = 105 \cr & {\text{or, }}{n^2} - n = 210 \cr & {\text{or, }}{n^2} - n - 210 = 0 \cr & {\text{or, }}n = 15,\, - 14 \cr} $$ But, we cannot take negative value of n So, n = 15 i.e. number of persons in the party = 15

One digit positive numbers = 5 Two digit positive numbers = 25 Three digit positive numbers = 100 4 digit positive numbers = 300 5 digit positive numbers = 600 Six digit positive numbers = 600 Total positive numbers, = 5 + 25 + 100 + 300 + 600 + 600 = 1630

The number of matches in first round,= 6 C 2 + 6 C 2 Number of matches in next round, = 6 C 2 Number of matches in semifinals, = 4 C 2 Total number of matches, = 6 C 2 + 6 C 2 + 6 C 2 + 4 C 2 + 2 = 53

Total number of ways = 7! × 2!

Step 1: Arrange the letters in alphabetical order The word KUBER consists of the letters: B, E, K, R, U (sorted order) Step 2: Count words before "KUBER" Words starting with 'B' = 4! = 24 Words starting with 'E' = 4! = 24 Words starting with 'K' and second letter 'B' = 3! = 6 Words starting with 'K' and second letter 'E' = 3! = 6 Words starting with 'K' and second letter 'R' = 3! = 6 Step 3: Find the position of "KUBER" After 24 (B) + 24 (E) + 6 (KB) + 6 (KE) + 6 (KR) = 66 words, the next word is KUBER Rank of "KUBER" = 67 Final Answer: Option A (67)

There are 9 non-zero digits to arrange themselves at 4 different position. Each letter can be arrange at different position in 9 different ways. So, required number of ways, = 9 × 9 × 9 × 9 = 9 4

5 students can be seated out of 10 students in 10 C 5 ways Remaining 5 will be seated in, = 5 C 5 ways Students of each row can be arranged as, = 5! × 5! ways Two sets of paper can be arranged themselves in, = 2! ways Thus, Total arrangement, = 10 C 5 × 5! × 5! × 2 = 7257600

Let the 3 boys be B 1 , B 2 , B 3 and 4 prizes be P 1 , P 2 , P 3 and P 4 Now B 1 is eligible to receive any of the 4 available prizes (so 4 ways) B 2 will receive prize from rest 3 available prizes(so 3 ways) B 3 will receive his prize from the rest 2 prizes available(so 2 ways) So total ways would be: 4 × 3 × 2 × 1 = 24 Ways Hence, the 4 prizes can be distributed in 24 ways

Total number of ways in which 8064 can be resolved as the product of two factors is 24 as below: (1,8064), (2,4032), (3,2688), (4,2016), (6,1344), (7,1152), (8,1008), (9,896), (12,672), (14,576), (16,504), (18,448), (21,884), (24,336), (28,288), (32,252), (36,224), (42,192), (48,168), (56,144), (63,128), (68,126), (72,112), (84,96)

The order of each letter in the dictionary is ABLORU. Now, with A in the beginning, the remaining letters can be permuted in 5! ways. Similarly, with B in the beginning, the remaining letters can be permuted in 5! ways. With L in the beginning, the first word will be LABORU, the second will be LABOUR. Hence, the rank of the word LABOUR is, 5! + 5! + 2 = 120 + 120 + 2 = 242 Note: 5! = 5 × 4 × 3 × 2 × 1 = 120

Two tallest boys can be arranged in 2! ways, rest 18 can be arranged in 18! ways. Girls can be arranged in 6! ways. Total number of ways in which all the students can be arranged, = 2! × 18! × 6! = 18! × 1440 Note: N! = N × (N - 1) × (N - 2) × . . . . × 1 So, 18! = 18 × 17 × 16 × 15 . . . . . . . . × 1

n P 3 = n × (n–1) × (n–2) (n+1) P 3 = (n+1) × n × (n–1) Now, 5 × n × (n–1) × (n–2) = 4 × (n+1) × n × (n–1) Or, 5(n−2) = 4(n+1) Or, 5n − 10 = 4n + 4 Or, 5n − 4n = 4 + 10 Hence, n = 14 Note: n P r = $$\frac{{{\text{n}}!}}{{\left( {{\text{n}} - {\text{r}}} \right)!}}$$

Number of words which have at least one letter replaced, = Total number of words - total number of words in which no letter is repeated. = 10 5 – 16 P 5 = 100000 − 30240 = 69760

Given there are 12 numbered chairs, such that chairs numbered 1 to 8 should be occupied. 1, 2, 3, 4, 5, 6, 7, 8 , 9, 10, 11, 12 The various combinations of chairs that ensure that no two men are sitting together are listed. (1, 3, 5, __ ), The fourth chair can be 5,6,10,11 or 12, hence 5 ways. (1, 4, 8, __ ), The fourth chair can be 6,10,11 or 12 hence 4 ways. (1, 5, 8, __ ), the fourth chair can be 10,11 or 12 hence 3 ways. (1, 6, 8, __ ), the fourth chair can be 10,11 or 12 hence 3 ways. (1, 8, 10, 12) is also one of the combinations. Hence, 16 such combinations exist. In case of each these combinations we can make the four men inter arrange in 4! ways. Hence, the required result =16 × 4! = 384

As A and T should occupy the first and last position, the first and last position can be filled in only one following way. A _ _ _ _ T. The remaining 4 positions can be filled in 4! Ways by the remaining words (S, C, E, N, T). Hence by rearranging the letters of the word ASCENT we can form, 1 x 4! = 4! Words.

n P r = n C r × r! 6 P r = 15 × r! 360 = 15 × r! r! = $$\frac{{360}}{{15}}$$ = 24 r! = 4 × 3 × 2 × 1 ⇒ r! = 4! Therefore, r = 4

Each ring may be worn in any of 4 fingers. So, each ring may be worn in 4 different ways. ∴ 6 rings may be worm in (4×4×4×4×4×4) = 4 6 = 4096 ways.

A triangle is formed by joining any three non-collinear points in pairs. There are 7 non-collinear points. The number of triangles formed, $$\eqalign{ & { = ^7}{{\text{C}}_3} \cr & = \frac{{7 \times \left( {7 - 1} \right) \times \left( {7 - 2} \right)}}{{3!}} \cr & = \frac{{7 \times 6 \times 5}}{{3 \times 2 \times 1}} \cr & = 7 \times 5 \cr & = 35 \cr} $$

To committee can be formed in the following ways, (1 lady + 4 gents) or (2 ladies + 3 gents) or (3 ladies + 2 gents) or (4 ladies + 1 gents) or (5 ladies + 0 gents). Total number of possible arrangements, = ( 4 C 1 × 6 C 4 ) + ( 4 C 2 × 6 C 3 ) + ( 4 C 3 × 6 C 2 ) + ( 4 C 4 × 6 C 1 ) = 60 + 120 + 60 + 6 = 246

At least 1 question from each section is compulsory, so from the 1 st section the candidate can attempt 1 or 2 or 3 or 4 questions. In each section each question can be dealt with in 2 ways, i.e. either he attempts it or leaves it. So far 4 question there are 2 × 2 × 2 × 2 ways to attempt. As he has to attempt at least 1 question, the total number of ways in which he can attempt questions from 1st section is 2 4 - 1 Similarly for the 2 nd section there are 2 5 - 1 ways in which he can attempt and for the 3 rd section there are 2 6 - 1 ways. The ways in which the attempts one or more questions in any section is independent of the number of ways in which he attempts one or more questions from the other sections. Thus, total number of ways in which he can attempt questions in that paper: = (2 4 - 1)(2 5 - 1)(2 6 - 1) = 15 × 31 × 63 = 29295

We are selecting three different posts here, so order matters. Thus, total ways of selecting a President, VP and Water-boy from a group of 10 people would be 10 P 3

Let there were x teams participating in the games, then total number of matches, = n C 2 = 153 On solving we get, n = −17 and n = 18 It cannot be negative so n = 18

Six balls can be selected in the following ways, One red ball and 5 blue balls Or Two red balls and 4 blue balls. Total number of ways, = ( 3 C 1 × 7 C 5 ) + ( 3 C 2 × 4 C 7 ) = 63 + 105 = 168

Required number of ways, = 3 C 2 × 4! × 4! = 1728

In order to reach (5,6) covering the shortest distance at the same time the man has to make 5 horizontal and 6 vertical steps. The number of ways in which these steps can be taken is given by, $$\eqalign{ & = \frac{{11!}}{{5! \times 6!}} \cr & = 462 \cr} $$

Since, all the points are equally spaced; hence the area of all the convex pentagons will be same.

5 questions can be selected in the following ways, 2 question from first group and 3 question from second group Or 3 question from first group and 2 question from second group. = ( 4 C 2 × 4 C 3 ) + ( 3 C 4 × 4 C 2 ) = 24 + 24 = 48

Let total number of persons present in the party be x, Then, $$\eqalign{ & \frac{{x \times \left( {x - 1} \right)}}{2} = 105 \cr & x = 15 \cr} $$

A pentagon has 5 sides. We obtain the diagonals by joining the vertices in pairs. Total number of sides and diagonals, = 5 C 2 = $$\frac{{5 \times 4}}{{2 \times 1}}$$ = 5 × 2 = 10 This includes its 5 sides also. ⇒ Diagonals = 10 – 5 = 5 Hence, the number of diagonals = 10 – 5 = 5