PROBLEMS

**Permutation and combination problems :**

Problems on permutation and combination play a key role in quantitative aptitude test. It is bit difficult to score marks in competitive exams without knowing the shortcuts related to permutation and combination.

**Permutations :**

The ways of arranging or selecting smaller or equal number of persons or objects from a group of persons or collection of objects with due regard being paid to the order of arrangement or selection are called permutations.

**Combinations : **

The number of ways in which smaller or equal number of things are arranged or selected from a collection of things where the order of selection or arrangement is not important are called combinations.

Selection is made. Beyond selection, order or arrangement is important. |
Selection is made. But arrangement or order is not important |

**Permutations : **

nPr = n!/(n-r)!

**Combinations :**

nCr = n!/r!(n-r)!

**Circular Permutations :**

Case (i) :

Both clockwise and anti clockwise rotations are considered. (Hint : Every person has the same two neighbors) Then, the formula for circular permutations is

(n-1)!

Case (ii) :

Either clockwise or anti clockwise rotation is considered, not both. (Hint: No person has the same two neighbors) Then, the formula for circular permutations is

(n-1)!

1. nPr = n(n-1)(n-2)....to "r" terms.

Example : 7P3 = 7x6x5 = 210

2. nCr = [n(n-1)(n-2)...to "r" terms]/r!

Example : 7P3 = [7x6x5]/[3x2x1] = 35

3. nCr = nCn-r

(we will use this property only when we want to reduce the value of "r")

Example : 25P22 = 25P3

**4. nP1 = n 5. nC1 = n 6. nP0 = 1 7. nC0 = 1 8. nPn = n! **

**(No. of permutations of n things taken all at a time) 9. nCn = 1 **

(Explanation : nCn = nCn-n = nC0 = 1)

**10. No. of Permutations of n things taken all at a time, ****when two particular things always come together is**

**= (n-1)!.2! **

**11. No. of Permutations of n things taken all at a time, when two particular things always do not come together is **

**= n!-(n-1)!.2!**

**12. The value of 0! = 1 **

**Multiplication rule : **

**There are two things, one can be done in "m" number of ways and the second can be done in "n" number of ways, Then the total number of ways of doing both the things is **

**= m x n**

**This rule is called multiplication rule. **

AND ===> Multiplication

**Addition rule :**

**There are two things, one can be done in "m" number of ways and the second can be done in "n" number of ways, Then the total number of ways of doing either the first one or second one, not both is **

**= m + n**

**This rule is called addition rule. **

OR ===> Addition

**Example 1 : **

Compute the sum of all 4 digit numbers which can be formed with the digits 1, 3, 5, 7, if each digit is used only once in each arrangement.

Solution :

No. of numbers can be formed with the digits 1,3,5 and 7 is

4! = 4 x 3 x 2 x 1 = 24

No. of digits given = 4 (they are 1,3,5 and 7)

Number of times, each digit of (1, 3, 5, 7) will repeat in 1's place, 10's place, 100's place and 1000's place is

= (No.of numbers formed) / (No.of digits given)

= 24/4

= 6

So, each of the given digits will occur six times in each of the place.

Sum of the digits in 1000's place is

= (1 + 3 + 5 + 7) x 6 = 16 x 6 = 96

Similar is the case in 100's place, 10's place and 1's place

The sum will be

aaaaaaaaaaaaaaa 96 x 1000 = 96000 aaaaaaaaaaaaaaaa aaaaaaaaaaaaaaa 96 x 100 = 9600 aaaaaaaaaaaaaaaa aaaaaaaaaaaaaaa 96 x 10 = 960 aaaaaaaaaaaaaaaa aaaaaaaaaaaaaaa 96 x 1 = 96 aaaaaaaaaaaaaaaa aaaaaaaaaaaaaaa ------------------------- aaaaaaaaaaaaaaaa aaaaaaaaaaaaaaa 106656 aaaaaaaaaaaaaaaa aaaaaaaaaaaaaa -------------------------- aaaaaaaaaaaaaaa

Hence, the sum of 4 digits numbers is 106656

Compute the sum of all 4 digit numbers which can be formed with the digits 1, 3, 5, 7, if each digit is used only once in each arrangement.

**Example 2 :**

The number of 4 digit numbers greater than 5000 can be formed out of the digits 3, 4, 5, 6 and 7 (no digit is repeated). Find the number of such numbers.

**Solution : **

Given digits are 3, 4, 5, 6 and 7 and we form 4 digit numbers greater than 5000.

1000's place ---> 3 choices (one of the digits of 5,6,7 is placed)

100's place ---> 4 choices (Having 4 out of 5 digits)

10's place ---> 3 choices (Having 3 out of 5 digits)

1's place ---> 2 choices (Having 2 out of 5 digits)

No. of numbers formed = 3 x 4 x 3 x 2 = 72

Hence, no.of numbers greater than 5000 can be formed is 72

**Example 3 :**

**A examination paper with 10 questions consists of 6 questions in Algebra and 4 questions in geometry. At least one question is to be attempted from each section. In how many ways can this be done ? **

**Solution :**

**We have "2" alternatives for each question. That is, either we may attempt or we may not attempt. Therefore, no. of ways to attempt six questions in Algebra is **

= 2^{6}

(But it includes the way of not attempting all the questions)

So, no. of ways to attempt at least one question in Algebra is

**= 2 ^{6 }- 1 **

**Similarly, no.of ways to attempt atleast one question in Geometry is**

** = 2 ^{4 }- 1**

**Total no.ways for both the sections is **

**= (2 ^{6 }- 1)(2^{4 }- 1) **

**= 945 **

Hence, the no. of ways of attempting atleast one question from each section is 945.

After having gone through the stuff given above, we hope that the students would have understood "Permutation and combination problems"

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