SOLUTION: How many values greater than 7000 can be formed using the digits 3, 5, 7, 8, 9 without repetition?

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Question 1209361: How many values greater than 7000 can be formed using the digits 3, 5, 7, 8, 9 without repetition?
Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3816)   (Show Source): You can put this solution on YOUR website!

Consider four-digit numbers of the form ABCD where A,B,C,D represent digits chosen from {3,5,7,8,9} with no repeats allowed.
The choices for A could be 7, 8 or 9 to ensure that the number exceeds 7000.
So we have 3 choices for slot A.
Whatever is selected for A, we have 4 choices for B, 3 for C, and 2 for D.
It yields 3*4*3*2 = 72 different ways to create the four-digit numbers.

Now consider the five-digit case. Numbers will look like ABCDE
Where A through E are chosen from {3,5,7,8,9} and no repeats are allowed.
There are 5*4*3*2*1 = 120 possible five-digit values larger than 7000.
Any permutation of these 5 digits will be larger than 7000 simply because any five-digit number is larger than any four-digit number.
We cannot form six-digit numbers or larger since we only have 5 digits to work with and repeats aren't allowed.

We found there are 72 ways to form the four-digit numbers, and 120 ways to form the five-digit numbers, such that whatever formed is larger than 7000.

Therefore,
72+120 = 192 is the final answer.
Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.
How many values greater than 7000 can be formed using the digits 3, 5, 7, 8, 9 without repetition?
~~~~~~~~~~~~~~~~~~~~~~~ 

There are two major options: the number can be any 4-digit, starting from any of the digits 7, 8, 9;

                       and   the number can be any 5-digit number.


In the first major option, there are 3 options for the 1st digit (7, 8 or 9);

                                     4 remaining options for the 2nd digit;

                                     3 remaining options for the 3rd digit;

                                     2 remaining options for the 4th digit;

                                     1 remaining option  for the 5th digit.


So, in the first major option, there are 3*4*3*2*1 = 72 possible different 4-digit numbers/values.


In the second major option, there are 5! = 120 possible different 5-digit numbers/values.


In all, there are 120 + 72 = 192 possible different numbers/values, satisfying imposed conditions.    ANSWER

Solved.



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