Permutation, Combination and Inclusion and Exclusion

 

Permutation, Combination and

Inclusion and Exclusion

M. Velrajan

1. How many five digit numbers can be formed with the digits 0, 1, 2, 4, 6 and 8?

(a) 800                            (b) 600 (c) 500 (d) 400

In a five digit number the first digit should not be 0, so it can be selected in 5 ways. 

If the digits are not repeating, in the second digit choice 0 has to be included with

the remaining 4 given digits, so that the second one can also be selected in 5 ways and

the third, fourth and fifth digits can be selected in 4, 3 and 2 ways.

The number of 5 digit numbers is 5 × 5 × 4 × 3 × 2 = 600. 

In this we assumed that the digits are not repeating. 

If the digits can be repeated then each of the second, third, fourth and fifth digit can

be selected in 6 ways, so the number of 5 digits numbers is 5 × 6 × 6 × 6 × 6 = 6480


2. How many 3 digit numbers can be generated from 1, 2, 3, 4, 5, 6, 7, 8, 9 such that the digits are in the ascending order?     (1) 80  (2) 81   (3) 83      (4) 84 

We list the possibilities of first, second and third digits

First    second      third

7              8              9                   1

 

6             7               8 or 9

6             8               9                   2+1 = 3 (note that 2 + previous)


5             6               7 or 8 or 9

5             7               8 or 9

5             8               9                   3 +2+1 = 6  (note that 3 + previous)    


4                                                   4+3+2+1 = 10 (4 + previous)

3                                                   5+10 = 15 (5 + previous)

2                                                   6+15 = 21                

1                                                   7+21 = 28

Total numbers = 1 + 3 + 6 + 10 + 15 + 21 + 28 = 84


3. There are 5 students P, Q, R, S and T. In how many ways can they sit in a

Q and R do not sit together?     (a)  120            (b) 48 (c) 66 (d) 72

The possible sitting places of Q and R are : 

Q               R                        

1               3 or 4 or 5          3 ways    

2               4 or 5                  2

3               5                         1

And Q and R can be interchanged

So, for Q and R there are 2 × ( 3 + 2 + 1) = 12 ways.

And for each of these ways the 3 others can sit in 3 × 2 × 1 = 6 ways. 

Hence the total ways = 12 × 6 = 72

  

4. There are 20 students in a class. If 8 of them are girls and a team of 2 boys and 2 girls is

to be selected from the class for its anniversary, in how many ways can it be done?

(a) 3696               (b) 1848 (c) 924 (d) 1396

Boys = 20 - 8 = 12.

2 boys can be selected in 12C2 = (12 ×  11)/2 = 66 ways

2 girls can be selected in 8C2 = (8 × 7)/2 = 28 ways

2 boys and 2 girls in 66 × 28 = 1848 ways        

Unit digit of 6 × 8 is 8 and in the answer choices only 1848 is with unit digit 8, so not necessary to multiply 66 and 28 completely.


5. There are 240 balls and n number of boxes. The balls are to be placed in the boxes such that the first

box should contain 4 balls more than the second box, the second box should contain 4 balls more than

the third box, and so on. Which one of the following cannot be the possible value of n?

(a) 4                   (b) 5             (c) 6 (d) 7

Suppose nth box has x balls. Then in n-1 has x+4, n-2 has x + 2 × 4,  . . . , box 2 has x + (n-2) × 4

and first box has x + (n-1) × 4 balls.

Hence 240 = x + x+4 + . . . + x+(n-1)×4 = nx + (1 + 2 + . . . +(n-1)) × 4 = nx + 2n(n-1).

Hence 240 is a multiple of n. 

240 is a multiple of 4, 5 and 6. But 240 is not a multiple of 7.  Ans. (d)


6. On a railway route between two places A and B, there are ten stations on the way. If four new stations

are to be added, how many types of new tickets will be required if each ticket is issued for a one way

journey?   (a) 14 (b) 48     (c) 96          (d) 108

Total existing stations including A and B = 12

After adding 4 new stations, total stations = 12 + 4 = 16

For any two stations one ticket for onward and one for the return journey.   

Types of tickets for existing stations = 2(12C2)

Types of tickets for after adding the 4 stations = 2(16C2)

Types of new tickets required = 2(16C2 - 12C2 ) = 2(120 - 66) = 108


7. In an examination there are three subjects A, B and C. A student has to pass in each subject. 20%

students failed in A, 22% students failed in B and 16% failed in C. The total number of students

passing the whole examination lies between 

(a) 42% and 84 %     (b) 42% and 78%   (c) 58% and 78%        (d) 58% and 84%      


Maximum 22% failed in B and minimum 16% failed in C.

The number of students passed in the whole examination is maximum if all the students who

failed in C also failed in B and all who failed in A also failed in B.

The maximum of the students passed in all = 100 - 22 = 78% 

The number of students passed in all is minimum if the students failed in each of the subjects

passed in the other 2 subjects.  

The minimum of students passed in all = 100 - (20 + 22 + 16) 

                                                                   = 100 - 58 = 42%

Ans. (b)


8. In an examination 30% of total students failed in English, 40% of students failed in Hindi and 20%

in both.   Find the percentage of students who passed in both the subjects.

(a)  50%                    (b)20%                           (c)10%                        (d)60%

         


         

A - English B - Hindi

Passed in A 70%, in B 60% and in at least one (A or B) = 80%. 

Passed in both A and B = 70 + 60 - 80 = 50%(since at least one is included in both passed in A and

in B, in A + B it is included two times, so subtracted it one time).


9. In an examination 10,000 students appeared. The result revealed the number of students who

have: Passed in all five subjects = 5583, Passed in three subjects only  = 1400, Passed in two

subjects only  = 1200, Passed in one subject only  = 735, failed in English only = 75, failed in

Physics only = 145, failed in Chemistry only = 140, failed in Mathematics only = 200, failed in

Bio-science only = 157 The number of students passed in at least four subjects is : 

(1) 6900 (2) 7300 (3) 7900 (4) 6300

Passed in at  least 4 = Passed in all 5 + Passed in 4 subjects only

Passed in 4 subjects only = failed in only one subject = 75 + 145 + 140 + 200 + 157 = 717

Passed in at  least 4 = 5583 + 717 = 6300

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