We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website.
Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see
cookie policy and privacy policy.
DECLINE COOKIES

How many 5-digit numbers have at least one 2 or one 3 among their digits?

ANotSmartPerson Oct 18, 2018

#1**+2 **

There are 99,999 - 10,000 + 1 = 90,000 five digit numbers

This equals 90,000 less the integers that contain no 2's or 3's

We have 7 ways to pick the leading digit [ since 0 cannot lead ] and 8 ways to pick each of the other 4 digits

So 7 * 8^4 = 28,762

So 90,000 - 28,762 = 61328 integers

CPhill Oct 18, 2018

#3**0 **

Of all the numbers between 10,000 and 99,999 =90,000 numbers, there are 300 2s in every 1,000 numbers. From this 300 will subtract repeats of which there are 10 per 1,000, namely:22, 122, 222, 322, 422, 522, 622, 722, 822, 922. That leaves us with 300 - 10 =290 numbers with 2s in them. So, **290 x 90,000/1,000 =26,100 **

In the 4th position, we have 1,000 2s x 9 =**9,000 2s.** Such as:12,000, 22,000, 32,000, 42,000, 52,000, 62,000, 72,000, 82,000, 92,000.

In the 5th position, we have** 10,000 2s**, namely:20,000 to 29,999. So, in total, we have:

**26,100 + 9,000 + 10,000 =45,100 numbers with 2s between 10,000 and 99,999.** This is in accord with the following computer count of ALL digits between 10,000 and 99,999, which adds 900 repeated(or 10 x 90,000/1,000) 2s to the count:

**(0=36 000, 1=46 000, 2=46 000, 3=46 000, 4=46 000, 5=46 000, 6=46 000, 7=46 000, 8=46 000, 9=46 000) The above applies exactly the same to 3s =45,100.**

Guest Oct 19, 2018