Question

The number of 4 digit numbers greater than 8000 that can be formed with two 5’s, three 8’s and four 9’s is
A. 51
B. 50
C. 52
D. 49

Answer 1

Hint: The 4 digit numbers must begin with 8 or 9, for to be greater than 8000. Then find the number of 4 digit numbers beginning with 8 and the number of 4 digit numbers beginning with 9. Exclude the numbers that are not possible with the given digits.

Complete step-by-step answer:
We are provided with two 5’s, three 8’s and four 9’s. We are asked to find the number of 4 digit numbers greater than 8000 that can be formed with these nine numbers.
Since the 4 digit number must be greater than 8000, the number should begin with a number greater than or equal to 8 (i.e. the digit at the ten-thousandth place must be greater than or equal to 8).
Therefore, all the numbers must begin with either 8 or 9.
Let us first find the number of 4 digit numbers that can be formed with 8 at the ten-thousandth place.
8 _ _ _
Now, we are left with two 5’s, two 8’s and four 9’s and we have to arrange them in three places.
This is similar to the case where we have to arrange ‘n’ different objects in ‘r’ places with repetition of the digits allowed.
The number of ways to arrange ‘n’ different objects in ‘r’ places with repetition of the digits allowed is equal to $ {{n}^{r}} $ .
Here, $ n=3 $ and $ r=3 $
Therefore, the number of ways 3 different digits can be arranged in 3 places with repetition is $ {{3}^{3}}=27 $ .
However, there are only two 5’s and two 8’s that can be used. Hence, the numbers 8555 and 8888 are not possible. So, these two cases must be excluded. .
This means that the number of 4 digit numbers that can be formed with 8 in the beginning is $ 27-2=25 $ .
Now, let us find the number of 4 digit numbers formed with 9 in the beginning.
9 _ _ _
We have two 5’s, three 8’s and three 9’s and we have to arrange them in three places.This is similar to above case, i.e. the number of ways 3 different digits can be arranged in 3 places with repetition is $ {{3}^{3}}=27 $ .
However, the number 9555 must be excluded since we only have two 5’s.
Hence, the number of 4 digit numbers that can be formed with 9 in the beginning is $ 27-1=26 $ .
Finally, the total number of 4 digit numbers that can be formed with the given digits is $ 25+26=51 $ .
Hence, the correct option is A.
So, the correct answer is “Option A”.

Note: This method used in the solution is one of the ways to find the answer.
Here, we can also each possible separately.
First, the beginning digit must be either 8 or 9. So choose 8 first.
Find the number of ways to arrange three distinct digits in three places. Then find the number of ways to arrange two similar and one different digits in the three places. Lastly, find the number of the ways to arrange all three similar numbers.
After this, use the same procedure for digit 9.