How many 4-digit numbers either begin or end with a 1?

Answer 1

Yes, there are 900 that begin with 1, and yes, there are 900 that end with 1. But there are NOT 1800 that either begin OR end with 1, because some of those -- exactly 100, in fact -- both begin AND end with 1, and you shouldn't count those twice.

SO the correct answer is 1700.

Answer 2

First, keep in mind there are 10 possibilities for a digit (0 - 9).

Begins with a 1:
1 choice for the first digit
10 choices for the second digit
10 choices for the third digit
10 choices for the fourth digit
Total: 1 * 10 * 10 * 10 = 1000

Ends with a 1:
9 choices for the first digit (first digit cannot be 0)
10 choices for the second digit
10 choices for the third digit
1 choice for the fourth digit
Total: 1 * 9 * 10 * 10 = 900

Begin and ends with a 1:
1 choice for the first digit
10 choices for the second digit
10 choices for the third digit
1 choice for the fourth digit
Total: 1 * 10 * 10 * 1 = 100


Let B be the # of 4-digit numbers that begin with a 1
Let E be the # of 4-digit numbers that end with a 1

We want BUE
Using the inclusion/exclusion principle:
BUE = B + E - B∩E
= 1000 + 900 - 100
= 1800

Answer is 1800.

Answer 3

I am assuming that you want the numbers that begin with 1 or end with 1, but do not both begin and end with 1.

1 * 10 * 10 * 9 begin with one.

9 * 10 * 10 *1 end with one.

Final answer 1800. (not!!!)

++++++++++++++++++++++++++++++++++++++++++++++++

Scott D has got this right.

We have counted numbers that have a zero in the thousands place in this estimation. These are not 4 digit numbers. To reiterate his point:

This is correct for those that begin with one:

1 * 10 * 10 * 9 = 900 begin with 1 and don't end with 1.

The numbers that end in 1 can only have 8 possible digits (2-9) in the thousands place:

8 * 10 * 10 * 1 = 800 end with 1 and don't begin with 1 or zero.

900 + 800 = 1700 (correct)

Another point of confusion here is some people are counting the numbers that both begin and end in one. I don't do that because in my opinion the "either/or" language implies not both. If you are counting the numbers with both a 1 in the one's place and a 1 in the thousand's place it is 1800.

Answer 4

your question says EITHER begins or ends with a 1, so im gonna take it that it cant be both (ie cant start and end with a 1)



so, 4 digit numbers only starting with a 1:

1 * 10 * 10 * 9 = 900
...................^
....(cant end with a 1 as well)



and, 4 digit numbers only ending with a 1:

8 * 10 * 10 * 1 = 800
^
(cant start with a 1 or a 0)



therefore, total number of 4 digit numbers that start either with a 1 or end with a 1:

900 +800 = 1700

Answer 5

I can't believe how many people are screwing up this easy question, some of them under the delusion that mathematical "or"s are commonly exclusive.

1000 4-digit numbers begin in 1
900 4-digit numbers end in 1

If you just add 1000 to 900, you're double-counting the 100 4-digit numbers that both begin and end in 1.

So your answer is 1800.

Answer 6

4 digit numbers

end with 1
1st digit = 1 to 9 (9 choices)
2nd and 3rd = 0-9 (10 choices)
last digit = is 1 (no choice)

start with 1
1st digit = is 1 (no choice)
2nd, 3rd and last = 0-9 (10 choices)

start AND end with 1
1st digit = is 1 (no choice)
2nd and 3rd = 0-9 (10 choices)
last digit = is 1 (no choice)


4-digit numbers either begin or end with a 1
= n(end) + n(start) -2(start n end)
= 9*10*10*1 + 1*10*10*10 - 2[1*10*10*1]
= 900 + 1000 - 200
= 1700

Answer 7

ha gotta use permutations and cominations!