Permutations And Combinations

Let the 3-digit number be ABC, where C is at the units place, B at the tens place and A at the hundreds place.

(i) When repetition is allowed:

The number of digits possible at C is 5. As repetition is allowed, the number of digits possible at B and A is also 5 at each.

Therefore, The total number possible 3-digit numbers =5 × 5 × 5 =125

(ii) When repetition is not allowed:

The number of digits possible at C is 5. Let’s suppose one of 5 digits occupies place C, now as the repletion is not allowed, the possible digits for place B are 4 and similarly there are only 3 possible digits for place A.

Therefore, The total number of possible 3-digit numbers=5 × 4 × 3=60

Let the 3-digit number be ABC, where C is at the unit’s place, B at the tens place and A at the hundreds place.

As the number has to even, the digits possible at C are 2 or 4 or 6. That is number of possible digits at C is 3. Now, as the repetition is allowed, the digits possible at B is 6 (any of the 6 is okay). Similarly, at A, also, the number of digits possible is 6.

Therefore, The total number possible 3 digit numbers =6 × 6 × 3 = 108.

Let the 4 digit code be 1234.

At the first place, the number of letters possible is 10. Let’s suppose any 1 of the ten occupies place 1. Now, as the repetition is not allowed, the number of letters possible at place 2 is 9. Now at 1 and 2, any 2 of the 10 alphabets have been taken. The number of alphabets left for place 3 is 8 and similarly the number of alphabets possible at 4 is 7.

The total number of 4 letter codes=10 × 9 × 8 × 7=5040.

Let the five-digit number be ABCDE. Given that first 2 digits of each number is 67. Therefore, the number is 67CDE.

As the repetition is not allowed and 6 and 7 are already taken, the digits available for place C are 0,1,2,3,4,5,8,9. The number of possible digits at place C is 8. Suppose one of them is taken at C, now the digits possible at place D is 7. And similarly, at E the possible digits are 6.

∴The total five-digit numbers with given conditions=8 × 7 × 6=336.

The possible outcomes after a coin toss are head and tail.

The number of possible outcomes at each coin toss is 2.

∴The total number of possible outcomes after 3 times=2 × 2 × 2=8.

The signal requires 2 flags.

The number of flags possible for upper flag is 5.

Now as one of the flag is taken, the number of flags remaining for lower flag in the signal is 4.

The number of ways in which signal can be given= 5 × 4 =20.

(i) 8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 40320

(ii) 4!-3! = (4 × 3!)-3! = 3!(4-1) = 3 × 2 × 1 × 3 = 18

Computing left hand side:

3!+4! = (3 × 2 × 1) + (4 × 3 × 2 × 1) = 6+24 = 30

Computing the right-hand side:

7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040

We observe that the left-hand side is not equal to the right-hand side.

Expanding all the factorials:

Solving left hand side:

⇒

Equating left hand side to right hand side:

⇒

⇒

⇒

(i) Putting the value of n and r:

⇒

(ii) Putting the value of n and r:

⇒

Total no. of digits possible for choosing = 9

No. of places for which a digit has to be taken = 3

As there is no repetition allowed;

⇒ No. of permutations =

In questions like these we need to fill the places that are to be occupied.

To find: Four digit number (digits does not repeat)

Now we will have 4 places where 4 digits are to be put.

So,

At thousand's place = There are 9 ways as 0 cannot be at thousand's place = 9 ways

At hundredth's place = There are 9 digits to be filled as 1 digit is already taken = 9 ways

At ten's place = There are now 8 digits to be filled as 2 digits are already taken = 8 ways

At unit's place = There are 7 digits that can be filled = 7 ways

Total Number of ways to fill the four places = 9 × 9 × 8 × 7 = 4536 ways.



So a total of 4536 four digit numbers can be there with no digits repeated.

Even number means that last digit should be even,

No. of possible digits at one’s place = 3 (2, 4 and 6)

⇒ No. of permutations=

One of digit is taken at one’s place, Number of possible digits available = 5

⇒ No. of permutations=

Therefore, total number of permutations =3 × 20=60.

Total no. of digits possible for choosing =5

No. of places for which a digit has to be taken =4

As there is no repetition allowed;

⇒ No. of permutations =

The number will be even when 2 and 4 are at one’s place.

The possibility of (2,4) at one’s place =

Total number of even number = 120 × 0.4 = 48.

Total no. of people in committee = 8

No. of positions to be filled = 2

⇒ No. of permutations =

⇒ n=9.

(i)

⇒

⇒

⇒

⇒ 42 -13r+r²=12

⇒ r²-13r+30=0

⇒ r²-10r-3r+30=0

⇒ r(r-10)-3(r-10)=0

⇒ (r-3)(r-10)=0

r= 3 or r=10

But r=10 is rejected, as in , r cannot be greater than 5.

Therefore, r=3.

(ii)

⇒

⇒

⇒

⇒ 42 -13r+r²=6

⇒ r²-13r+36=0

⇒ r²-9r-4r+36=0

⇒ r(r-9)-4(r-9)=0

⇒ (r-4)(r-9)=0

r= 4 or r=9

But r=9 is rejected, as in , r cannot be greater than 5.

Therefore, r=4.

Total number of different letters in EQUATION = 8

Number of letters to be used to form a word =8

⇒ No. of permutations =

Total number of letters in MONDAY =6

(i) No. of letters to be used =4

⇒ No. of permutations =

(ii) No. of letters to be used =6

⇒ No. of permutations =

(iii) No. of vowels in MONDAY = 2 (O and A)

⇒ No. of permutations in vowel =

Now, remaining places =5

Remaining letters to be used =5

⇒ No. of permutations =

Therefore, total number of permutations = 2 × 120 =240.

Total number of letters in MISSISSIPPI =11

Letter Number of occurrence

⇒ Number of permutations =

We take that 4 I’s come together, and they are treated as 1 letter,

∴ Total number of letters=11-4+1=8

⇒ Number of permutations =

Therefore, total number of permutations where four I’s don’t come together = 34650-840=33810

Total number of letters in PERMUTATIONS =12

Only repeated letter is T ; 2times

(i) First and last letter of the word are fixed as P and S respectively.

Number of letters remaining =12-2=10

⇒ No. of permutations =

(ii) No. of vowels in PERMUTATIONS = 5 (E,U,A,I,O)

Now, we consider all the vowels together as one.

Number of permutations of vowels =

Now total number of letters = 12-5+1=8

⇒ No. of permutations =

Therefore, total number of permutations = 120× 20160=2419200

(iii) Number of places are as 123456789101112

There should always be 4 letters between P and S.

Possible places of P and S are 1 and 6, 2and 7, 3 and 8, 4 and 9, 5 and 10, 6 and 11, 7 and 12

Possible ways =7,

Also, P and S can be interchanged,

No. of permutations =2× 7 =14

Remaining 10 places can be filled with 10 remaining letters,

∴ No. of permutations =

Therefore, total number of permutations = 14 × 1814400 =25401600.

Given: ⁿC₈ = ⁿC₂

We know that if ⁿC_r = ⁿC_p then either r = p or r = n – p

Here ⁿC₈ = ⁿC₂

⇒ 8 = n – 2

⇒ n = 10

Now,

∴ ⁿC₂= ¹⁰C₂ = (∵ ⁿC_r = )

⇒¹⁰C₂ =

(i)

Given: =12 : 1

⇒

⇒

⇒

⇒

⇒ 4× (2n—1) = 12×(n-2)

⇒ 8n – 4 = 12n – 24

⇒ 12n – 8n = 24 – 4

⇒ 4n = 20

∴ n = 5

(ii)

Given: =11 : 1

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒ 4× (2n—1) = 11×(n-2)

⇒ 8n – 4 = 11n – 22

⇒ 11n – 8n = 22 – 4

⇒ 3n = 18

∴ n = 6

Given: 21 points on a circle

We know that we require two points on the circle to draw a chord

∴ Number of chords is:

⇒ ²¹C₂=

∴ Total number of chords can be drawn are 210

Given: 5 boys and 4 girls are in total

We can select 3 boys from 5 boys in ⁵C₃ ways

Similarly, we can select 3 boys from 54 girls in ⁴C₃ ways

∴ No. of ways a team of 3 boys and 3 girls can be selected is ⁵C₃ × ⁴C₃

⇒ ⁵C₃ × ⁴C₃ =

⇒ ⁵C₃ × ⁴C₃ = 10 × 4 = 40

∴ No. of ways a team of 3 boys and 3 girls can be selected is ⁵C₃ × ⁴C₃ = 40 ways

Given: 6 red balls, 5 white balls and 5 blue balls

We can select 3 red balls from 6 red balls in ⁶C₃ ways

Similarly, We can select 3 white balls from 5 white balls in ⁵C₃ ways

Similarly, We can select 3 blue balls from 5 blue balls in ⁵C₃ ways

∴ No. of ways of selecting 9 balls is ⁶C₃ ×⁵C₃ × ⁵C₃

⇒ ⁶C₃ ×⁵C₃ × ⁵C₃ =

⇒ ⁶C₃ ×⁵C₃ × ⁵C₃ =

∴ Number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour is ⁶C₃ ×⁵C₃ × ⁵C₃ = 2000

Given: a deck of 52 cards

There are 4 Ace cards in a deck of 52 cards.

According to question, we need to select 1 Ace card out the 4 Ace cards

∴ No. Of ways to select 1 Ace from 4 Ace cards is ⁴C₁

⇒ More 4 cards are to be selected now from 48 cards (52 cards – 4 Ace cards)

∴ No. Of ways to select 4 cards from 48 cards is ⁴⁸C₄

∴ Total number of ways = ⁴C₁ × ⁴⁸C₄

⇒ ⁴C₁ × ⁴⁸C₄ =

⇒ ⁴C₁ × ⁴⁸C₄ =

∴ Number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination 778320.

Given: 17 players in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers

There are 5 players how can bowl, and we require 4 bowlers in a team of 11

∴ No. Of ways in which bowlers can be selected are: ⁵C₄

Now other players left are: 17 – 5(bowlers) = 12

Since we need 11 players in a team and already 4 bowlers are selected, we need to select 7 more players from 12.

∴ No. Of ways we can select these players are: ¹²C₇

∴ Total number of combinations possible are : ⁵C₄ × ¹²C₇

⇒ ⁵C₄ × ¹²C₇ =

⇒⁵C₄ × ¹²C₇ =

∴ Number of ways we can select a team of 11 players where 4 players are bowlers from 17 players are: 3960

Given: A bag contains 5 black and 6 red balls

Number of ways we can select 2 black balls from 5 black balls are ⁵C₂

Number of ways we can select 3 red balls from 6 red balls are ⁶C₃

∴ Number of ways 2 black and 3 red balls can be selected are: ⁵C₂× ⁶C₃

∴ ⁵C₂× ⁶C₃ =

⇒ ⁵C₂× ⁶C₃ =

∴ Number of ways in which 2 black and 3 red balls can be selected from 5 black and 6 red balls are : 200

Given: 9 courses are available and 2 specific courses are compulsory for every student

Here 2 courses are compulsory out of 9 courses, so a student need to select 5—2=3 courses

∴ number of ways in which 3 ways can be selected from 9—2(compulsory courses) = 7 are ⁷C₃

∴ ⁷C₃ =

⇒ ⁷C₃ =

∴ Number of ways a student selects 5 courses from 9 courses where 2 specific courses are compulsory are: 35

The word DAUGHTER has 3 vowels A, E, U and 5 consonants D, G, H, T and R.

The three vowels can be chosen in ³C₂ as only two vowels are to be chosen.

Similarly, the five consonants can be chosen in⁵C₃ ways.

∴ Number of choosing 2 vowels and 5 consonants would be ³C₂ ×⁵C₃

= 30

∴ Total number of ways of is 30

Each of these 5 letters can be arranged in 5 ways to form different words = ⁵P₅

Total number of words formed would be = 30 × 120 = 3600

in the word EQUATION there are 5 vowels (A, E, I, O, U) and 3 consonants (Q, T, N)

The numbers of ways in which 5 vowels can be arranged are ⁵C₅

………………1

Similarly, the numbers of ways in which 3 consonants can be arranged are ³P₃

……………..2

There are two ways in which vowels and consonants can appear together

(AEIOU)(QTN) or (QTN)(AEIOU)

∴ the total number of ways in which vowel and consonant can appear together are:

2 × ⁵C₅ × ³C₃

∴ 2 × 120 × 6 = 1440

(i) exactly 3 girls

Total numbers of girls are 4

Out of which 3 are to be chosen

∴ Number of ways in which choice would be made = ⁴C₃

Numbers of boys are 9 out of which 4 are to be chosen which is given by ⁹C₄

Total ways of forming the committee with exactly three girls

= ⁴C₃ × ⁹C₄

=

(ii) at least 3 girls

There are two possibilities of making committee choosing at least 3 girls

There are 3 girls and 4 boys or there are 4 girls and 3 boys

Choosing three girls we have done in (i)

Choosing four girls and 3 boys would be done in ⁴C₄ ways

And choosing 3 boys would be done in ⁹C₃

Total ways = ⁴C₄ ×⁹C₃

Total numbers of ways of making the committee are

504 + 84 = 588

(iii) at most 3 girls

In this case the numbers of possibilities are

0 girl and 7 boys

1 girl and 6 boys

2 girls and 5 boys

3 girls and 4 boys

Number of ways to choose 0 girl and 7 boys = ⁴C₀ × ⁹C₇

Number of ways of choosing 1 girl and 6 boys = ⁴C₁ × ⁹C₆

Number of ways of choosing 2 girls and 5 boys = ⁴C₂ × ⁹C₅

Number of choosing 3 girls and 4 boys has been done in (1)

= 504

Total number of ways in which committee can have at most 3 girls are = 36 + 336 + 756 + 504 = 1632

In dictionary words are listed alphabetically, so to find the words

Listed before E should start with letter either A, B, C or D

But the word EXAMINATION doesn`t have B, C or D

Hence the words should start with letter A

The remaining 10 places are to be filled by the remaining letters of the word EXAMINATION which are E, X, A, M, 2N, T, 2I, 0

Since the letters are repeating the formula used would be

Where n is remaining number of letters

p₁ and p₂ are number of times the repeated terms occurs.

The number of words in the list before the word starting with E

= words starting with letter A = 907200

The number is divisible by 10 if the unit place has 0 in it.

The 6-digit number is to be formed out of which unit place is fixed as 0

The remaining 5 places can be filled by 1, 3, 5, 7 and 9

Here n = 5

And the numbers of choice available are 5

so, the total ways in which the rest the places can be filled are ⁵P₅

There are 5 vowels and 21 consonants in English alphabets.

Choosing two vowels out of 5 would be done in ⁵C₂ ways

Choosing 2 consonants out of 21 can be done in ²¹C₂ ways

The total number of ways selecting 2 vowels and 2 consonants

= ⁵C₂ × ²¹C₂

Each of these four letters can be arranged in four ways ⁴P₄

Total numbers of words that can be formed are

24 × 2100 = 50400

The student can choose 3 questions from part I and 5 from part II

Or

4 questions from part I and 4 from part II

5 questions from part 1 and 3 from part II

3 questions from part I and 5 from part II can be chosen in

= ⁵C₃ × ⁷C₅

4 questions from part I and 4 from part II can be chosen in

= ⁵C₄ × ⁷C₄

5 questions from part 1 and 3 from part II can be chosen in

= ⁵C₅ × ⁷C₃

Now the total number of ways in which a student can choose the questions are = 210 + 175 + 35 = 420

A deck of cards has 4 kings.

The numbers of remaining cards are 52.

Ways of selecting a king from the deck = ⁴C₁

Ways of selecting the remaining 4 cards from 48 cards= ⁴⁸C₄

Total number of selecting the 5 cards having one king always

= ⁴C₁ × ⁴⁸C₄

There are total 9 people

Women occupies even places that means they will be sitting on 2^nd , 4^th , 6^th and 8^th place where as men will be sitting on 1^st, 3^rd, 5^th,7^th and 9^th place.

4 women can sit in four places and ways they can be seated= ⁴P₄

5 men can occupy 5 seats in 5 ways

The numbers of ways in which these can be seated = ⁵P₅

The total numbers of sitting arrangements possible are

24 × 120 = 2880

There are 2 options

Total numbers of ways in which students can be chosen are

=170544 + 646646

= 817190

In the given word ASSASSINATION, there are 4 ‘S’. Since all the 4 ‘S’ have to be arranged together so let as take them as one unit.

The remaining letters are= 3 ‘A’, 2 ‘I’, 2 ‘N’, T

The number of letters to be arranged are 9 (including 4 ‘S’)

Using the formula where n is number of terms and p₁, p₂ p₃ are the number of times the repeating letters repeat themselves.

Here p₁= 3, p₂= 2, p₃ = 2

Putting the values in formula we get