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[Ruka]

Cho các số từ $0$ đến $9$ . Lập các số tự nhiên có $5$ chữ số khác nhau . Tính xác suất chọn được $1$ số mà không có $2$ chữ số chẵn đứng kề nhau?

[Ruka]

KGM: $|\Omega| = 9 . A_9^4$

We will analyse $4$ cases : 

Case $1$ : There are $5$ odd numbers : 

Pick $5$ odd number and put them together to form a number has $5!$(ways)

 

Case $2$ : There are $4$ odd numbers and $1$ even number

Pick $4$ odd nums we have $\displaystyle\binom{5}{4}$

Choosing randomly $1$ even num has $\displaystyle\binom{5}{1}$

mix them we have $5!$ ways

$\to$ We can create $\displaystyle\binom{5}{4} . \displaystyle\binom{5}{1} . 5!$(nums)

With $0$ in the head of the number we can create $\displaystyle\binom{5}{4}  . 4!$(nums)

In altogether : $\displaystyle\binom{5}{4} . \displaystyle\binom{5}{1} . 5! - \displaystyle\binom{5}{4}  . 4! = 2880$

 

Case $3$ : There are $2$ even numbers

_ With no $0$ 

Choose and priortize the first $3$ odd nums : $A_5^3$

There are $2$ spaces between the odd nums and $2$ outer space, place $2$ even nums there we have $A_4^2$

$\to$ There're : $A_5^3 . A_4^2 = 720(nums)$

_ With a $0$

Choose and priortize the first $3$ odd nums : $A_5^3$

Choose other even number(not 0) $\to$ $C_4^1$ ways

Choose $2$ places for $2$ even number $\to C_4^2$ ways

To sum up, there're $A_5^3.C_4^1.C_4^2$(nums)

Let $0$ be the head of the num , we can create $A_5^3.C_4^1.C_3^1$

In altogether, we can create : $A_5^3.C_4^1.C_4^2  -  A_5^3.C_4^1.C_3^1 = 720$ (nums)

This case has $720 . 2 = 1440$(nums)

Case $4$ : There are $3$ even numbers

Applying the same method we can create $480 + 480 = 960$ (nums)

Call A  ... 

So we have $P(A) = \dfrac{|A|}{|\Omega|} = \dfrac{25}{126}$

Bài viết đã được chỉnh sửa nội dung bởi Ruka: 03-02-2023 - 20:50