Permutation and Combination

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Directions: Each of the questions given below is followed by series of options among which you have to choose the best one. You can also review your choices by clicking into answer section.


1. Find the number of triangles in a heptagon.

A. 28

B. 35

C. 21

D. 14


Answer

                  Ans: Option B

       Explanation:

       There are 7 vertices in a heptagon. No. of ways of selecting 3 vertices out of 7 = 7C3 = 35


2. In how many different ways can the letters of the word 'MATHEMATICS' be arranged such that the vowels must always come together?

A. 9800

B. 100020

C. 120960            

D. 140020


Answer

                  Ans: Option C

       Explanation:

      The word 'MATHEMATICS' has 11 letters. It has the vowels 'A','E','A','I' in it and these 4 vowels must always come together.       Hence these 4 vowels can be grouped and considered as a single letter. That is,                     MTHMTCS (AEAI).

      Hence we can assume total letters as 8. But in these 8 letters, 'M' occurs 2 times, 'T' occurs 2 times but rest of the letters is       different.

      Hence, number of ways to arrange these letters = 8! (2!)(2!) =8×7×6×5×4×3×2×1(2×1)(2×1)=10080

      In the 4 vowels (AEAI), 'A' occurs 2 times and rest of the vowels is different.

      Number of ways to arrange these vowels among themselves = 4! 2!=4×3×2×12×1=12

      Hence, required number of ways = 10080 x 12 = 120960


3. 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


Answer

                  Ans: Option C

       Explanation:

       'LOGARITHMS' contains 10 different letters.

       Required number of words= Number of arrangements of 10 letters, taking 4 at a time.

                = 10P4

                = (10 x 9 x 8 x 7)

                = 5040.


4. In how many different ways can the letters of the word 'DETAIL' be arranged such that the vowels must occupy only the odd positions?

A. None of these            

B. 64

C. 120   

D. 36


Answer

                  Ans: Option D

       Explanation:

       The word 'DETAIL' has 6 letters which has 3 vowels (EAI) and 3 consonants (DTL). The 3 vowels (EAI) must occupy only the odd positions.

       Let's mark the positions as (1) (2) (3) (4) (5) (6).

       Now, the 3 vowels should only occupy the 3 positions marked as (1),(3) and (5) in any order.

      Hence, number of ways to arrange these vowels = 3P3

       = 3! = 3 x 2 x 1 = 6

     Now we have 3 consonants (DTL) which can be arranged in the remaining 3 positions in any order.

      Hence, number of ways to arrange these consonants = 3P3

      = 3! = 3 x 2 x 1 = 6

     Total number of ways

     = number of ways to arrange the vowels x number of ways to arrange the consonants

      = 6 x 6 = 36


5. 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


Answer

                  Ans: Option C

       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.


6. 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. None of these

C. 1800 

D. 3600


Answer

                  Ans: Option C

       Explanation:

       5 subjects can be arranged in 6 periods in 6P5 ways.

       Remaining 1 period can be arranged in 5P1 ways.

       Two subjects are alike in each of the arrangement. So we need to divide by 2! to avoid over-counting.

       Total number of arrangements = (6P5 x 5P1)/2! = 1800

      Alternatively this can be derived using the following approach.

      5 subjects can be selected in 5C5 ways.

      Remaining 1 subject can be selected in 5C1 ways.

     These 6 subjects can be arranged themselves in 6! ways.

      Since two subjects are same, we need to divide by 2!

      Total number of arrangements = (5C5 × 5C1 × 6!)/2! = 1800


7. Find the number of diagonals in an octagon.

A. 8

B. 16

C. 28

D. 20


Answer

                  Ans: Option D

       Explanation:

       An octagon has 8 vertices. No. of ways of connecting 2 vertices = 8C2 = 8*7/2 = 28 Subtracting the 8 sides, the no. of diagonals = 28-8 = 20


8. In how many ways can 10 software engineers and 10 civil engineers be seated in a row so that they are positioned alternatively?

A. 2 × (10!)2       

B. 2 × 10! × 11!

C. 10! × 11!         

D. (10!)2


Answer

                  Ans: Option 

       Explanation:

       The 10 civil engineers can be arranged in a row in

       10P10 = 10! Ways —(A.

       Now we need to arrange software engineers such that software engineers and civil engineers are seated alternatively. i.e., we can arrange 10 software engineers either in the 10 positions marked as                                A,B,C,D,E,F,G,H,I,J or in the 10 positions marked as B,C,D,E,F,G,H,I,J,K, as shown below:

       10 software engineers can be arranged in the 10 positions marked as A,B,C,D,E,F,G,H,I,J in 10P10 = 10! Ways

       10 software engineers can be arranged in the 10 positions marked as B,C,D,E,F,G,H,I,J,K in 10P10 = 10! Ways

        10 software engineers can be arranged in the 10 positions marked as A,B,C,D,E,F,G,H,I,J or in the 10 positions marked as B,C,D,E,F,G,H,I,J,K in 10! + 10! = 2 × 10! Ways —(B.

        From (A) and (B)

       The required number of ways = 10! × (2 × 10!) = 2 × (10!)2


9. 8 friends go for a trip in 2 cars. Each car can accommodate a maximum of 4 persons. In how many ways can they travel?

A. 160

B. 140

C. 168

D. 70

Answer: Option D


Answer

                  Ans: Option D

       Explanation:

       Required no. of ways = No. of ways of selecting 4 persons for the first car (since the remaining four will have to travel in the second car) = 8C4 = 8!/(4!.4!) = 70 


10. 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


Answer

                  Ans: Option C

       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) as:

        5

       Hence the total count of 1 digit numbers that can be formed using the 5 digits (2,4,5,6,8) (repetition alloweD. = 5 —(A)

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

       Since repetition is allowed, any of the 5 digits(2,4,5,6,8) can be placed in unit place and tens place as:

        5              5

        Hence the total count of 2 digit numbers that can be formed using the 5 digits (2,4,5,6,8) (repetition alloweD. = 52 —(B) 

        Count of 3 digit numbers that can be formed using the 5 digits (2,4,5,6,8) (repetition allowed)

        Since repetition is allowed, any of the 5 digits (2,4,5,6,8) can be placed in unit place , tens place and hundreds place.

         5              5              5

        Hence the total count of 3 digit numbers that can be formed using the 5 digits (2,4,5,6,8) (repetition alloweD. = 53 —(C)

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

        Since repetition is allowed, any of the 5 digits (2,4,5,6,8) can be placed in unit place, tens place, hundreds place and thousands place as:

        5              5              5              5

       Hence the total count of 4 digit numbers that can be formed using the 5 digits (2,4,5,6,8) (repetition alloweD. = 54 —(D)

       From (A) (B) (C) and (D)

      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

      =5(54–1)5−1 [∵ Reference: Sum of first n terms in a geometric progression (G.P.) ]

      =5(625–1)4=5(624)4=5×156=780


11. Find the total number of 9 digit numbers which have all the digits different.

A. 9 * 9 !

B. 10!

C. 9!

D. None of these


Answer

                  Ans: Option A

       Explanation:

       10P9 – 9P8 = 9 * 9!


12. From a group of 8 women and 6 men, a committee consisting of 3 men and 3 women is to be formed. In how many ways can the committee be formed if two of the women refuse to serve together?

A. 1020

B. 1000

C. 712   

D. 896


Answer

                  Ans: Option B

       Explanation:

       Let the women be X and Y who refuses to serve together.

       Let's find out the number of ways in which the committee can be formed by excluding both X and Y.

       We excluded both X and Y. Hence we need to select 3 women from 6 women i.e. (8-2) and 3 men from 6 men. The number of ways in which this can be done = 6C3 × 6C3 —(A)

       Now let's find out the number of ways in which the committee can be formed where exactly one woman from X and Y will be present. i.e., we need to select one woman from two women (X and Y), remaining 2                women from 6 women i.e. (8-2) and 3 men from 6 men. The number of ways in which this can be done

       = 2C1 × 6C2 × 6C3 — (B)

       From (A) and (B), the number of ways in which a committee be formed if two of the women refuses to serve together

        = 6C3 × 6C3 + 2C1 × 6C2 × 6C3

        = 6C3 (6C3 + 2C1 × 6C2 )

         = (6×5×43×2×1)[(6×5×43×2×1)+2(6×52×1)]

         = (8×7)[4+(4×3)]=20[20+30]=20×50=1000


13. 4 buses run between Bhopal and Gwalior. If a man goes from Gwalior to Bhopal by a bus and comes back to Gwalior by another bus, then the total possible ways are

A. 12

B. 4

C. 16

D. 8


Answer

                  Ans: Option A

       Explanation:

       Since the man can go in 4 ways and can back in 3 ways. Therefore total number of ways is ways.


14. A box contains 12 different black balls, 7 different red balls and 6 different blue balls. In how many ways can the balls be selected?

A. 728   

B. 225 – 1

C. 225   

D. 727


Answer

                  Ans: Option B

       Explanation:

       Total number of combinations is the total number of ways of selecting one or more than one things from n distinct things. i.e., we can select 1 or 2 or 3 or … or n items at a time.

       Total number of combinations = nC1 + nC2 + … + nCn = 2n – 1

       It is explicitly stated that 12 black balls are different, 7 red balls are different  and 6 blue balls are different. Hence there are 25(=12+ 7+ 6) different balls.

       We can select one ball from 25 balls, two balls from 25 balls, … 25 balls from 25 balls. Hence, required number of ways

       = Number of ways in which 1 ball can be selected from 25 distinct balls + Number of ways in which 2 balls can be selected from 25 distinct balls + Number of ways in which 3 balls can be selected from 25 distinct          balls. . .+ Number of ways in which 25 balls can be selected from 25 distinct balls

       = 25C1 + 25C2 + … + 25C25 = 225 – 1


15. The number of ways in which 6 rings can be worn on the four fingers of one hand is.

A. 4⁶

B. 6⁴

C. 6C4

D. None of these


Answer

                  Ans: Option A

       Explanation:

       No explanation


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