Permutations and Combinations

A $ \times $ B will have 2 $ \times $ 4 = 8 elements.

The number of subsets having atleast 3 elements

= ⁸C₃ + ⁸C₄ + ⁸C₅ + ⁸C₆ + ⁸C₇ + ⁸C₈

= 2⁸ – (⁸C₀ + ⁸C₁ + ⁸C₂) = 256 – 1 – 8 – 28 = 219

For alike n objects, number of ways we can select zero or more objects = n + 1 and number of ways we can select at least one object = n

Given 10 identical white balls, 9 identical green balls and 7 identical black balls.

To find number of ways for selecting atleast one ball.

Number of ways to choose zero or more white balls = (10 + 1) [since, all white balls are mutually identical]

Number of ways to choose zero or more green balls = (9 + 1) [since, all green balls are mutually identical]

Number of ways to choose zero or more black balls = (7 + 1) [since, all black balls are mutually identical]

Hence, number of ways to choose zero or more balls of any colour = (10 + 1) (9 + 1) (7 + 1)

Also, number of ways to choose a total of zero balls = 1

Hence, the number, if ways to choose at least one ball (irrespective of any colour) = (10 + 1) (9 + 1) (7 + 1) $-$ 1 = 879

Let XA, X_B, X_C and X_D represent number of balls present in box A, B, C and D respectively.

As no box can be empty so,

X_A $\ge$ 1, X_B $\ge$ 1, X_C $\ge$ 1 and X_D $\ge$ 1

$\Rightarrow$ X_A $-$ 1 $\ge$ 0, $\Rightarrow$ X_B $-$ 1 $\ge$ 0, $\Rightarrow$ X_C $-$ 1 $\ge$ 0 and $\Rightarrow$ X_D $-$ 1 $\ge$ 0

t_A $\ge$ 0, t_B $\ge$ 0, t_C $\ge$ 0 and t_D $\ge$ 0

According to the question,

X_A + X_B + X_C + X_D = 10

$\Rightarrow$ (X_A $-$ 1) + (X_B $-$ 1) + (X_C $-$ 1) + (X_D $-$ 1) = 6

$\Rightarrow$ t_A + t_B + t_C + t_D = 6

Now question becomes, box A, B, C, and D can have none or one or more balls and total balls are 6

From formula we know, n things can be distributed among r people in ${}^{n + r - 1}{C_{r - 1}}$ ways where each people can have either 0 or more things.

$\therefore$ 6 balls can be distributed among 4 boxes in ${}^{6 + 4 - 1}{C_{4 - 1}} = {}^9{C_3}$ ways where each box can have either 0 or more balls.

Therefore, Statement 1 is correct. The number of ways of choosing any 3 places from 9 different places is ${}^9{C_3}$ ways. But Statement - 2 is not the correct explanation of Statement - 1.

We need 3 points to create a triangle. With 10 points number of triangle possible ${}^{10}{C_3}$

Here 6 points are on the same line so we can't make any triangle with those 6 points.

So subtract ${}^{6}{C_3}$.

$\therefore$ $N = {}^{10}{C_3} - {}^6{C_3}$

$ = {{10\,.\,9\,.\,8} \over {1\,.\,2\,.\,3}} - {{6\,.\,5\,.\,4} \over {1\,.\,2\,.\,3}}$

$ = 120 - 20$

$ = 100$

Thus number of ways $ = ({}^3{C_2}) \times ({}^9{C_2}) = 3 \times {{9 \times 8} \over 2} = 108$

From 6 different novels 4 novels can be chosen = ${}^6{C_4}$ ways

And from 4 different dictionaries 1 can be chosen = ${}^3{C_1}$ ways

$\therefore$ From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary can be chosen = ${}^6{C_4} \times {}^3{C_1}$ ways

Let 4 novels are N₁, N₂, N₃, N₄ and 1 dictionary is D₁.

Dictionary should be in the middle. So the arrangement will be like this

_ _ D₁ _ _

On those 4 blank places 4 novels N₁, N₂, N₃, N₄ can be placed. And 4 novels can be arrange $4!$ ways.

$\therefore$ Total no of ways = ${}^6{C_4} \times {}^3{C_1}$$ \times 4!$ = 1080

This problem is solved using gap method. As here no 'S' is adjacent to each other so we have to put them in the gap. So first write all the letters other than 'S' such a way that there is a gap between two letters.

Given word is MISSISSIPPI.

Here, I = 4 times, S = 4 times, P = 2 times, M = 1 time

_M_I_I_I_I_P_P_

Those seven letters M, I, I, I, I, P, P can be arranged in ${{7!} \over {4!2!}}$ ways

Those seven letters creates 8 gaps and we have to choose 4 gaps from those 8 gaps to put those four 'S' letters.

This can be done ${}^8{C_4}$ ways.

After placing those four 'S' letters we can arrange them in ${{4!} \over {4!}}$ ways.

Therefore, required number of words

$ = {{7!} \over {4!2!}} \times {}^8{C_4} \times {{4!} \over {4!}}$

$ = {{7\,.\,6!} \over {4!4!}} \times {}^8{C_4}$

$ = 7\,.\,{}^6{C_4}\,.\,{}^8{C_4}$

Note : n items can be distribute among p persons are ${}^{n + p - 1}{C_{p - 1}}$ ways.

Here n = 6 ice-cream

p = 5 types of ice-cream

Each ice-cream belongs to one of the 5 ice-cream type. So chosen 6 ice-crean can be divide into 5 types of ice-cream.

$ \therefore $ The number of different ways the child can buy the six ice-cream is = ${}^{6 + 5 - 1}{C_{5 - 1}}$ = ${}^{10}{C_4}$

$ \therefore $ Statement - 1 is false.

Number of different ways of arranging 6 A and 4 B's in a row

= ${{10!} \over {6!4!}} = {}^{10}{C_4}$

$ \therefore $ Statement - 2 is true.

The total number of ways is

${}^{12}{C_4} \times {}^{12 - 4}{C_4} \times {}^{12 - 4 - 4}{C_4} = {}^{12}{C_4} \times {}^8{C_4} \times {}^4{C_4} = {{12!} \over {{{(4!)}^3}}}$

A voter can give vote to either 1 candidate or 2 candidates or 3 candidates or 4 candidates.

Case 1 : When he give vote to only 1 candidate then no ways = ${}^{10}{C_1}$

Case 2 : When he give vote to 2 candidates then no ways = ${}^{10}{C_2}$

Case 3 : When he give vote to 3 candidates then no ways = ${}^{10}{C_3}$

Case 4 : When he give vote to 4 candidates then no ways = ${}^{10}{C_4}$
So, total no of ways he can give votes
= ${}^{10}{C_1} + {}^{10}{C_2} + {}^{10}{C_3} + {}^{10}{C_4}$
= 385

Note : Here we use addition rule as he can vote any one of those four rules. Whenever there is "or" choices, we use addition rule.

In the word ''GARDEN'', there are two vowels A and E present, and A should come always before E.

$\therefore\,\,\,$ Total no of ways = ${{6!} \over {2!}}$ = 360

Here A and E has fixed order that is why we divide by 2!.

To distribute n objects among p people where everyone should get atleast one object, then number of ways to distribute those n objects

= ${}^{n - 1}{C_{p - 1}}$

For this question, n = 8 and p = 3

$ \therefore $ Number of ways = ${}^{8 - 1}{C_{3 - 1}}$ = ${}^7{C_2}$ = 21

6 men can sit at the round table = $\left( {6 - 1} \right)! = 5!$ ways

Now at the round table among 6 men there are 6 empty places and 5 women can sit at those 6 empty positions.



So total no of ways 6 men and 5 women can dine at the round table

= $5!\, \times {}^6{C_5} \times 5!$

= $5!\, \times 6 \times 5!$

= $5!\, \times 6!$

Case 1 :

No of ways student can answer 10 questions = ${}^5{C_4} \times {}^8{C_6}$ = 140

Case 2 :

No of ways student can answer 10 questions = ${}^5{C_5} \times {}^8{C_5}$ = 56

$\therefore$ Total ways = 140 + 56 = 196

Arrange it this way,

$^n{C_{r + 1}} + 2.{}^n{C_r} + {}^n{C_{r - 1}}$

$ = {}^n{C_{r + 1}} + {}^n{C_r} + {}^n{C_r} + {}^n{C_{r - 1}}$

$\left[ \, \right.$ Now use the rule,

        $\left. {{}^n{C_r} + {}^n{C_{r - 1}} = {}^{n + 1}{C_r}} \right]$

$ = {}^{n + 1}{C_{r + 1}} + {}^{n + 1}Cr$

$ = {}^{n + 2}{C_{r + 1}}$

Note : For a number to be divisible by 3, the sum of digits should be divisible by 3.

Here given numbers are 0, 1, 2, 3, 4 and 5. Out of those 6 numbers possible sets of 5 numbers are (1, 2, 3, 4, 5) and (0, 1, 2, 4, 5) whose sum are divisible by 3.

Set 1 : Set is = (1, 2, 3, 4, 5). Sum of digits = 1 + 2 + 3 + 4 + 5 = 15 (Divisible by 3)



So total no of arrangement = 1$ \times $2$ \times $3$ \times $4$ \times $5 = 5!

Set 2 : Set is = (0, 1, 2, 4, 5). Sum of digits = 0 + 1 + 2 + 4 + 5 = 12 (Divisible by 3)

So total no of arrangement = 4$ \times $4$ \times $3$ \times $2$ \times $1 = 4.4!

$\therefore$ Total arrangement = 5! + 4.4! = 216

$\therefore$ Total no of ways = 5$ \times $6$ \times $6$ \times $${}^4{C_1}$ = 720

According to this question, any number between 1 to 100 should be divisible by 2 or 5 but not by 2$ \times $5 = 10.

Possible numbers between 1 to 100 divisible by 2 are 2, 4, 6, .... , 100

This is an A.P where first term = 2, last term = 100 and total terms = 50.

$ \therefore $ Sum of the numbers divisible by 2

= ${{50} \over 2}\left[ {2 + 100} \right]$

= 25$ \times $102

= 2550

Possible numbers between 1 to 100 divisible by 5 are 5, 10, 15, .... , 100

$ \therefore $ Sum of the numbers divisible by 5

= ${{20} \over 2}\left[ {5 + 100} \right]$

= 10$ \times $105

= 1050

And possible numbers between 1 to 100 divisible by 10 are 10, 20, 30, .... , 100

$ \therefore $ Sum of the numbers divisible by 10

= ${{10} \over 2}\left[ {10 + 100} \right]$

= 5$ \times $110

= 550

$ \therefore $ Required sum = 2550 + 1050 - 550 = 3050

There are 3 possible ways that we can make number greater than 1000 but less than 4000 using the digits 0, 1, 2, 3, 4 where repetition is allowed

Case 1 : First digit is 1 = 1 _ _ _

Possible numbers starting with 1 = 1$ \times $5$ \times $5$ \times $5 = 125

But this includes 1000 also which does not satisfy the given condition of being greater than 1000. Hence there will be 124 numbers having 1 in the first place.

Case 2 : First digit is 2 = 2 _ _ _

Possible numbers starting with 2 = 1$ \times $5$ \times $5$ \times $5 = 125

Case 3 : First digit is 3 = 3 _ _ _

Possible numbers starting with 3 = 1$ \times $5$ \times $5$ \times $5 = 125

Total possible numbers = 124 + 125 + 125 = 374