Permutations and Combinations

Start with

(1) $\overline{E}:\frac{6!}{2!}=360$  

(2)    $\overline{GE}:\frac{5!}{2!},$ $\overline{GN}:\frac{5!}{2!}$  

(3) GTE : 4!, GTN: 4!, GTT : 4!

(4) GTWENTY = 1

⇒ 360 + 60 + 60 + 24 + 24 + 24 + 1 = 553

x + 2y + 3z = 42

0    x + 2y = 42 ->22 cases

1    x + 2y = 39 ->19 cases

2    x + 2y = 36 ->19 cases

3    x + 2y = 33 ->17 cases

4    x + 2y = 30 ->16 cases

5    x + 2y = 27 ->14 cases

6    x + 2y = 24 ->13 cases

7    x + 2y = 21 ->11 cases

8    x + 2y = 18 ->10 cases

9    x + 2y = 15 ->8 cases

10  x + 2y =12 -> 7 cases

11  x + 2y = 9 -> 5 cases

12  x + 2y = 6 -> 4 cases

13  x + 2y = 3 -> 2 cases

14  x + 2y = 0 -> 1 cases.

Total ways to partition 5 into 4 parts are:

5 0 

4 1 0  $\frac{5!}{4!}=5$

3 2 0 $\frac{5!}{3!\cdot 2!}=10$

3 1 0 $\frac{5!}{2!2!2!}=15$

2 1 $\frac{5!}{2!*3!}=10$

51 Total way

After giving 2 apples to each child 15 apples left now 15 apples can be distributed in
^15+3–1C₂ = ¹⁷C₂ ways

  $=\frac{17*16}{2}=136$          

$=8*7*6*5*4+\frac{9*8}{2}$

= 6756

18 = 3² * 2

For G.C.D to be 3. no. of four digits should be only multiple of 3, but not multiple of 9 & also should not be even.

As we know no. of the form                         

9 k -> 1000

9 k + 1 -> 1000

9 k + 2 -> 1000

9 k + 3 -> 1000  -> Total no. = 2000

9 k + 4 -> 1000

9 k + 5 -> 1000

9 k + 6 ->1000

9 k + 7 -> 1000

9 k + 8 -> 1000

In which half will be even & half be odd so Required no. = 1000

Boys (10)            Girls (5)

  (3)                        (3)

B₁ & B₂ should not be selected together

Total number of ways    

= (56 + 56) * 10 = 1120

Fix the unit place, find the chances for the first three digits

unit digit as 1, total ways = 9.10²

unit digit as 2, total ways = 4.5²

unit digit as 3 total ways = 3.4²

unit digit as 4 total ways = 2.3²

unit digit as 5 total ways = 1.2²

unit digit as 6 total ways = 1.2²

unit digit as 7 total ways = 1.2²

unit digit as 8 total ways = 1.2²

unit digit as 9 total ways = 1.2²

Sum of digits

1 + 2 + 3 + 5 + 6 + 7 = 24

So, either 3 or 6 rejected at a time

Case 1 Last digit is 2

……….2

no. of cases = 

  Case 2 Last digit is 6

……….6

= 4! = 24

Total cases = 72

Factors of 36 = 2².3².1

Five-digit combinations can be 

(1, 2, 3, 3), (1, 4, 3, 1), (1, 9, 2, 1), (1, 4, 9, 11), (1, 2, 3, 6, 1), (1, 6, 1, 1)

i.e., total numbers  $\frac{5!5!}{2!2!}+\frac{5!}{2!2!}+\frac{5!}{2!2!}+\frac{5!}{3!}+\frac{5!}{2!}+\frac{5!}{3!2!}=\left(30*3\right)+20+60+10=180.$