Posted on

How many – part II

Q:  How many strings of four decimal digits
(a) Do not contains the same digit twice?
(b) Begin with an odd digit?
(c) Have exactly 3 digits that are 9?

A:  Back to the slot method… Four digits: _ _ _ _

(a) Do not contain the same digit twice (basically asking for no repeats).. So, how many choices are there for the first slot?  This questions is actually up in the air.  Can 0 occupy the first digit or not?  Usually, for example 0456 = 456 (which is a three-digit-number, not a four-digit-number)… This all depends on your interpretation, or your teacher’s interpretation!

I believe due to the wording of this problem that 0 can occupy the first slot.  Since we are looking for “strings” and not necessarily four-digit numbers.  The first slot may be 0-9 (10 choices):

10 _ _ _

Now, the second digit can be 0-9 (10 choices).. However, it cannot be the same as the first choice, so minus 1 choice = 9 choices.

10 9 _ _

The third digit can be 0-9 (10 choices), but minus 2 choices (cannot match slot 1 or slot 2) = 8 choices

10 9 8 _

Fourth slot only has 7 choices (same logic as above)

10 9 8 7

There are 10*9*8*7 total 4-digit numbers with distinct digits.

(b)  Begin with an odd digit?

Again, four digits: _ _ _ _

The first slot must be odd: 1, 3, 5, 7, 9 = 5 choices

5 _ _ _

The rest of the digits can be any choice 0-9 (10 choices)

5 10 10 10

5*10*10*10 total four-digit numbers that start with an odd digit!

(c)  Have exactly three digits that are 9?

Many ways to solve this… But, we will do it my way.. Ha, we always do it my way on here…

So, the number must look like one of the follow:

(1) 9 9 9 _ or

(2) 9 9 _ 9 or

(3) 9_ 9 9 or

(4) _ 9 9 9

The 9s above do not represent choices, but rather that the number 9 is locked in and must be 9, no other number.  So, in option (1) there are 9 choices for the final digit (the number can be 0-8).  In option (2) there are 9 choices for the blank (0-8 are your options).  Again for option (3): there are 9 choices for the blank.  And, option (4) has 9 choices for the blank too.

Now, this is an “or” problem.  You have option (1) or (2) or (3) or (4).  In counting, or means addition.

So, there are 9 + 9 + 9 + 9 = 36 total four-digit strings that have exactly 3 nines.

Leave a Reply