[combinatorics] permutation and combination (example of permutation and combination)

Reference blog :

• 【 Combinatorial mathematics 】 Basic counting principle ( The principle of addition | Multiplication principle )
• 【 Combinatorial mathematics 】 Examples of permutation and combination of sets ( array | Combine | Circular arrangement | binomial theorem )
• 【 Combinatorial mathematics 】 Permutation and combination ( Arrange and combine content summary | Select the question | Set arrangement | Set combination )

One 、 Examples of permutations 1 ( Combine | product rule | The law of addition )

---

The basic counting formula is The law of addition , product rule ;

from

1

1

1 ~

300

300

300 Take out at will

3

3

3 Number , So that the sum of these three numbers can be

3

3

3 to be divisible by , How many selection methods are there ?

Use classification ( product rule ) , Distribution ( The law of addition ) , Permutation and combination To solve the problem ;

Put the above

1

1

1 ~

300

300

300 Numbers , Divide by

3

3

3 The remainder of is divided into the following three categories :

• ① Divide $3$

  1 :

• ② Divide $3$

  2 :

• ③ Divide $3$

  0 :

Combination question :

• stay

  A

  A

  A Set any

  3

  3

  3 Number , The sum of the three numbers must be

  3

  3

  3 Multiple , Can times

  3

  3

  3 to be divisible by ; The selection methods are

  C

  (

  100

  ,

  3

  )

  C(100, 3)

  C(100,3) Kind of ;

• stay

  B

  B

  B Set any

  3

  3

  3 Number , The sum of the three numbers must be

  3

  3

  3 Multiple , Can times

  3

  3

  3 to be divisible by ; The selection methods are

  C

  (

  100

  ,

  3

  )

  C(100, 3)

  C(100,3) Kind of ;

• stay

  C

  C

  C Set any

  3

  3

  3 Number , The sum of the three numbers must be

  3

  3

  3 Multiple , Can times

  3

  3

  3 to be divisible by ; The selection methods are

  C

  (

  100

  ,

  3

  )

  C(100, 3)

  C(100,3) Kind of ;

product rule : stay

A

,

B

,

C

A,B,C

A,B,C Take a number from each set in , The sum of the three numbers is also

3

3

3 Multiple ,

• The first set takes $1$
• The second set takes $1$
• The third set takes $1$

All in all

10

0

3

100^3

1003 To plant and take ;

The final choice , Use the rule of addition :

3

C

(

100

,

3

)

+

10

0

3

=

1485100

3C(100, 3) + 100^3 = 1485100

3C(100,3)+1003=1485100

Two 、 Examples of permutations 2

---

1000

!

1000!

1000! At the end of

0

0

0 The number of ?

This value is calculated by multiplication , A very large , Basically impossible to calculate ;

List the factors : take

1000

!

1000!

1000! regard as

1000

×

999

×

998

×

⋯

×

2

×

1

1000 \times 999 \times 998 \times \cdots \times 2 \times 1

1000×999×998×⋯×2×1

Factor ;

Principle that : Among the above factors

1000

1000

1000 A factor , Will this

1000

1000

1000 Factorization , If there is

i

i

i individual

5

5

5 ,

j

j

j individual

2

2

2, be

i

i

i and

j

j

j The smaller value in

min

⁡

{

i

,

j

}

\min\{ i,j \}

min{ i,j} Namely

0

0

0 The number of ;

Above

1

1

1 ~

1000

1000

1000 this

1000

1000

1000 It is statistically decomposed from numbers

2

2

2 and

5

5

5 The number of

Statistics

2

2

2 The number of factors : Certainly more than 500 ;

• ① yes $2$

  500 individual

• ② yes $4$

  250 individual , Break it down

• ③ yes $16$

  62 individual , Break it down

• ④ yes $32$

  31 individual , Break it down

Statistics

5

5

5 The number of factors :

249

249

249 individual ;

• ① yes $5$

  200 individual , Statistics

• ② yes $25$

  40 individual , Break it down

• ③ yes $125$

  8 individual , Break it down

• ④ yes $625$

  1 individual , Break it down

It's broken down into

2

2

2 The number of

i

i

i It must be greater than

500

500

500 Number of numbers ;

It's broken down into

5

5

5 The number of

j

j

j The value is

249

249

249 individual ;

therefore

1000

!

1000!

1000! At the end of

0

0

0 The number of yes

249

249

249 individual ;