অগ্রণী ব্যাংক সিনিয়র অফিসার প্রস্তুতি: নিমিষেই করে ফেলুন কঠিন বিন্যাস ও সমাবেশ

09 Dec 2020

Permutations and Combinations

 

Arranging Objects

The number of ways of arranging n unlike objects in a line is n! (pronounced ‘n factorial’). n! = n × (n – 1) × (n – 2) ×…× 3 × 2 × 1

Example

How many different ways can the letters P, Q, R, S be arranged?

The answer is 4! = 24.

This is because there are four spaces to be filled: _, _, _, _

The first space can be filled by any one of the four letters. The second space can be filled by any of the remaining 3 letters. The third space can be filled by any of the 2 remaining letters and the final space must be filled by the one remaining letter. The total number of possible arrangements is therefore 4 × 3 × 2 × 1 = 4!

• The number of ways of arranging n objects, of which p of one type are alike, q of a second type are alike, r of a third type are alike, etc is:

n!        .
p! q! r! …

Example

In how many ways can the letters in the word: STATISTICS be arranged?

There are 3 S’s, 2 I’s and 3 T’s in this word, therefore, the number of ways of arranging the letters are:

    10!     =50 400
3! 2! 3!

Rings and Roundabouts

• The number of ways of arranging n unlike objects in a ring when clockwise and anticlockwise arrangements are different is (n – 1)!

When clockwise and anti-clockwise arrangements are the same, the number of ways is ½ (n – 1)!

Example

Ten people go to a party. How many different ways can they be seated?

Anti-clockwise and clockwise arrangements are the same. Therefore, the total number of ways is ½ (10-1)! = 181 440

Combinations

The number of ways of selecting r objects from n unlike objects is:

 

image: http://revisionworld.com/sites/revisionworld.com/files/imce/Permut4.gif

 

 

Example

There are 10 balls in a bag numbered from 1 to 10. Three balls are selected at random. How many different ways are there of selecting the three balls?

¹⁰C₃ =10!   =10 × 9 × 8                    = 120
           3!     (10 – 3)!3 × 2 × 1

 

Permutations

A permutation is an ordered arrangement.

• The number of ordered arrangements of r objects taken from n unlike objects is:

ⁿP_r =       n!       .
          (n – r)!

Example

In the Match of the Day’s goal of the month competition, you had to pick the top 3 goals out of 10. Since the order is important, it is the permutation formula which we use.

¹⁰P₃ =10!
            7!

= 720

There are therefore 720 different ways of picking the top three goals.

Probability

The above facts can be used to help solve problems in probability.

Example

In the National Lottery, 6 numbers are chosen from 49. You win if the 6 balls you pick match the six balls selected by the machine. What is the probability of winning the National Lottery?

The number of ways of choosing 6 numbers from 49 is ⁴⁹C₆ = 13 983 816 .

Therefore the probability of winning the lottery is 1/13983816 = 0.000 000 071 5 (3sf), which is about a 1 in 14 million chance.

 

Q.In how many different way the words ‘HOUSE’ can be arranged ? 
A. The word HOUSE contains 5 letters.
Required number of word using formula 5p5 = 5! = ( 1 x 2 x 3 x 4 x 5 ) = 120.

 

Q.In how many different way the word ‘CAROM’ can be arranged ?
A.The word CAROM contains 6 letters.
and two word are repeated So we divide the total word by repeated word.
So,6

 

Q.In how many different way the words ‘COMPUTER’ can be arranged ?
A. The word COMPUTER contains 7 letters.Here is no repeated word so,Required number of word using formula, ⁷p₇ = 7! / ( 7 – 7)! = ( 1 x 2 x 3 x 4 x 5 x 6 x 7 ) = 5040.

 

Permutation examples 1 with tricks
7! = 7 x 6 x 5 x 4 x 3 x 2 x 1 = 5040

 

Q.In how many different way the words ‘KOLKATA’ can be arranged ?
A. The word KOLKATA contains 7 letters.Here is repeated word occurred so we divided this by repeated word.Repeated is k two time and A two time,

Required number of word using formula ⁷p₇ / ²p₂ x ²p₂ = 7!/(7 – 7 )! / 2!(2 – 2)! x 2!(2 – 2)! = ( 1 x 2 x 3 x 4 x 5 x 6 x 7 ) = 1260.

 

Permutation examples 1 with tricks
7! / 2! x 2! = 7 x 6 x 5 x 4 x 3 x 2 x 1 / 2 x 1 x 2 x 1 = 1260.

 

Example  1 :
In how many several way the word KOLKATA can be prepared so that as vowel always come together?
Answer:
First of all we need to know how many vowel in this given word, here is three vowel that is O and A,A.
then we count the consonant that is four.Now we count the number of vowel as a single unit that is vowel O A,A count as single unit and add it with consonant so we have a four unit.
(4 consonant +three vowel as single unit ) = 5 x 3( three vowel)

5! x 3! = 1 x 2 x 3 x 4 x 5 x 1 x 2 x 3 = 720 ways we prepared the word.

Example 2 :
In how many ways can 6 apples be distributed among 5 boys , there being no restriction to the number of apples each boy may get ?
Answer :
Any of the 5 boys get each apple and this can be done in 5 ways
The number of  ways is = 56 = 15625 .

Example 3 :
In how many several way the word HOUSE can be prepared so that as vowel always come together?
Answer :
First of all we need to know how many vowel in this given word, here is three vowel that is E ,O and U.
then we count the consonant that is Two.Now we count the number of vowel as a single unit that is vowel E,O and U count as single unit and add it with consonant so we have a  three unit.
(2 consonant + three vowel as single unit ) = 3 x 3( three vowel)

3! x 3! = 1 x 2 x 3 x 1 x 2 x 3 = 36 ways we prepared the word.

Example 4:  In how many several ways the word CAME can be arranged,that the vowels not come together?
Answer : First of all we need to know how many vowel in this given word, here is two vowel that is A and E.

then we count the consonant that is Two.Now we count the number of vowel as a single unit that is vowel A and E count as single unit and add it with consonant so we have a three unit.
(2 consonant + two vowel as single unit ) = 3 x 2( two vowel)

Note: If we Subtract total number ways from within vowel ways than we get the not come to-gather ways,
So, ( 4! – 3! x 2! ) = 12.

Example 5:  In how many several ways the word HATE can be arranged, that the vowels not come together?
Answer :  First of all we need to know how many vowel in this given word, here is two vowel that is A and E.

then we count the consonant that is Two.Now we count the number of vowel as a single unit that is vowel A and E count as single unit and add it with consonant so we have a three unit.
(2 consonant + two vowel as single unit ) = 3 x 2( two vowel)

Note: If we Subtract total number ways from within vowel ways than we get the not come together ways,
So, ( 4! – 3! x 2! ) = 12

 

What is Combination ?
Each of the different groups or selections which can be formed by taking some or all of a number of objects, is called a combination. In combination we can select only one item at time it based on selection on choice.
Suppose Ball , Marbel, etc.

Type 1: Suppose we want to select two out of three girls A, B, C. Then possible selection are AB, BC, and CA.
Note: that AB and BA represent the same selection.

Type 2: Suppose We select all three girls Like A, B, C. Then the selection is ABC.
Note: BAC, ABC, CAB are the same choice of selection.

Type 3: All the combinations formed by a, b, c, taking two at a time are ab, bc, ca.

Type4: The only combinations that can be formed  of three letters a, b, c taken all at a time is abc.

Type5: Various person of 2 out of four person A, B, C, D are : AB,  AC, AD, BC, BD, CD.

Type6: Note that ab and ba are two different permutations but they represent the same combination.

Number of Combinations : The number of all combinations of n things, taken r at a time is :

ⁿ C _r = n! / r! ( n – r ) ! = n ( n – 1 ) ( n – 2 ) …….to r factors / r!.

Note: if n = r , ⁿC_r = 1 and ⁿ C ₀ = 1.

Important Rule:
ⁿC_r = ⁿ C _{(n – r )}. = 1.

Ex: ¹⁶ C ₁₃ = ¹⁶ C _{( 16 – 13 )} = ¹⁶ C _{3 = 16 x 15 x 14 x 13! / 3! = 16 x 15 x 14 / 3 x 2 x 1 = 560}

Find out value of ¹⁰ P <sub>3

=</sub> ¹⁰ C ₃= 10 x 9 x 8 / 3! = 10 x 9 x 8 / 3 x 2 x 1 = 120

Find the value of ⁵⁰ C ₅₀

= ⁵⁰ C ₅₀ = 1. [ n C n = 1]

 

More example: 

* In how many different ways can the letters of the word 'OPTICAL' be arranged so that the vowels always come together?

The word 'OPTICAL' contains 7 different letters.

When the vowels OIA are always together, they can be supposed to form one letter.

Then, we have to arrange the letters PTCL (OIA).

Now, 5 letters can be arranged in 5! = 120 ways.

The vowels (OIA) can be arranged among themselves in 3! = 6 ways.

Required number of ways = (120 x 6) = 720.

 

** In how many different ways can the letters of the word 'MATHEMATICS' be arranged so that the vowels always come together?

In the word 'MATHEMATICS', we treat the vowels AEAI as one letter.

Thus, we have MTHMTCS (AEAI).

Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different.

Number of ways of arranging these letters = 8!/{ (2!)(2!)}

= 10080.

Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.

Number of ways of arranging these letters

= 4!/2!= 12.

So, Required number of words = (10080 x 12) = 120960.

 

*** 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?

 

'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.

 

** In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?

 

Required number of ways = (7C5 x3C2) = (7C2 x 3C1)

= {(7 x 6)/( 2 x 1)} x 3 = 63.

 

** In how many different ways can the letters of the word 'DETAIL' be arranged in such a way that the vowels occupy only the odd positions?

 

There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants.

Let us mark these positions as under:

     (1) (2) (3) (4) (5) (6)

Now, 3 vowels can be placed at any of the three places out 4, marked 1, 3, 5.

Number of ways of arranging the vowels = 3P3 = 3! = 6.

Also, the 3 consonants can be arranged at the remaining 3 positions.

Number of ways of these arrangements = 3P3 = 3! = 6.

Total number of ways = (6 x 6) = 36.

 

* In how many different ways can the letters of the word 'DETAIL' be arranged in such a way that the vowels occupy only the odd positions?

There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants.

Let us mark these positions as under:

     (1) (2) (3) (4) (5) (6)

Now, 3 vowels can be placed at any of the three places out 4, marked 1, 3, 5.

Number of ways of arranging the vowels = 3P3 = 3! = 6.

Also, the 3 consonants can be arranged at the remaining 3 positions.

Number of ways of these arrangements = 3P3 = 3! = 6.

Total number of ways = (6 x 6) = 36.

 

*In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women?

Required number of ways = (8C5 x10C6)

= (8C3 x 10C4)

={(8 x 7 x 6) / ( 3 x 2 x 1)} x {(10 x 9 x 8 x 7) / (4 x 3 x 2 x 1)}

= 11760.

 

* How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?

 

Since each desired number is divisible by 5, so we must have 5 at the unit place.

So, there is 1 way of doing it.

The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9).

So, there are 5 ways of filling the tens place.

The hundreds place can now be filled by any of the remaining 4 digits.

So, there are 4 ways of filling it.

So, Required number of numbers = (1 x 5 x 4) = 20.

 

** In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?

We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys).

Required number of ways = (6C1 x4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4)

= (6C1 x 4C1) + (6C2 x 4C2) + (6C3x 4C1) + (6C2)

= (6 x 4) + {(6 x 5)/( 2 x 1) x (4 x 3)/( 2 x 1)} + [{(6 x 5 x 4)/(3 x 2 x 1)}x 4] + (6 x 5)/( 2 x 1)

= (24 + 90 + 80 + 15)

= 209.

*  In how many different ways can the letters of the word 'CORPORATION' be arranged so that the vowels always come together?

 

In the word 'CORPORATION', we treat the vowels OOAIO as one letter.

Thus, we have CRPRTN (OOAIO).

This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different.

Number of ways arranging these letters =7!/ 2!= 2520.

Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged

In 5!/ 3! = 20 ways.

Required number of ways = (2520 x 20) = 50400.

 

* From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?

 

We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).

Required number of ways = (7C3 x6C2) + (7C4 x 6C1) + (7C5)            

= (7 x 6 x 5)/( 3 x 2 x 1) x (6 x 5)/( 2 x 1) + (7C3 x 6C1) + (7C2)

= 525 + [{(7 x 6 x 5)/( 3 x 2 x 1)}x 6] + (7 x 6)/( 2 x 1)

= (525 + 210 + 21)

= 756.