Probability: Find the number of ways to form 4-digit numbers >3000

[Joe_1234]

Please help me solve this,

Given digits 2,2,3,3,3,4,4,4,4, how many distinct 4 digit numbers greater than 3000 can be formed?

Thank you

 

[MarkFL]

What have you tried, and where are you stuck?

 

[Joe_1234]

What have you tried, and where are you stuck?

2×3×3×3=54, is this correct sir? I am confused since 51 is the key answer

 

[MarkFL]

The first digit is either 3 or 4. So, let's consider each case separately:

(1) First digit is 3:

Then the rest of the numbers must come from the list: 2, 2, 3, 3, 4, 4, 4, 4
Therefore we may choose any 3-digit sequence except 222 and 333 for the rest of the digits. This shows there are:

\(\displaystyle N_1=3^3-2=25\)

numbers in this case.

(2) First digit is 4:

Then the rest of the numbers must come from the list 2, 2, 3, 3, 3, 4, 4, 4
Therefore we may choose any 3-digit sequence except 222 for the rest of the digits. This shows there are:

\(\displaystyle N_2=3^3-1=26\)

numbers in this case.

Hence, the total number is:

\(\displaystyle N=N_1+N_2=51\)

 

[Joe_1234]

The first digit is either 3 or 4. So, let's consider each case separately:

(1) First digit is 3:

Then the rest of the numbers must come from the list: 2, 2, 3, 3, 4, 4, 4, 4
Therefore we may choose any 3-digit sequence except 222 and 333 for the rest of the digits. This shows there are:

\(\displaystyle N_1=3^3-2=25\)

numbers in this case.

(2) First digit is 4:

Then the rest of the numbers must come from the list 2, 2, 3, 3, 3, 4, 4, 4
Therefore we may choose any 3-digit sequence except 222 for the rest of the digits. This shows there are:

\(\displaystyle N_2=3^3-1=26\)

numbers in this case.

Hence, the total number is:

\(\displaystyle N=N_1+N_2=51\)

Thank you sir