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

AI Thread Summary
To find the number of distinct 4-digit numbers greater than 3000 using the digits 2, 2, 3, 3, 3, 4, 4, 4, 4, the first digit must be either 3 or 4. If the first digit is 3, the remaining digits can be selected from 2, 2, 3, 3, 4, 4, 4, resulting in 25 valid combinations. If the first digit is 4, the remaining digits yield 26 valid combinations. Therefore, the total number of distinct 4-digit numbers greater than 3000 is 51. The calculations confirm the solution is accurate.
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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
 
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What have you tried, and where are you stuck?
 
MarkFL said:
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
 
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:

$$N_1=3^3-2=25$$

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:

$$N_2=3^3-1=26$$

numbers in this case.

Hence, the total number is:

$$N=N_1+N_2=51$$
 
MarkFL said:
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:

$$N_1=3^3-2=25$$

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:

$$N_2=3^3-1=26$$

numbers in this case.

Hence, the total number is:

$$N=N_1+N_2=51$$
Thank you sir
 
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