Permutations and Combinations

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