+0

# counting and probability help

0
270
4
+1315

How many three-digit whole numbers have at least one 7 or at least one 9 as digits?

No overcounting, so for example, 179 or 197 is counted ONCE and not TWICE.

Thanks!

Feb 2, 2022

#1
+676
+1

107, 109, 117, 119, 127, 129, 137, 139, 147, 149, 157, 159, 167, 169, 170, 171, 172, 173, 174, 175, 176, 177, 178, 179, 187, 189, 190, 191, 192, 192, 193, 194, 195, 196, 197, 198, 199 (37)

207, ... , 299 (37)

...

700, ... , 799 (100)

...

900, ... , 999 (100)

37 * 7 = 259

259 + 100 + 100 = 459

777 and 999 (- 2) and 171/191, 272/292, 373/393, 474/494, 575/595, 676/696, 797, 878/898, 979 (- 16)

(- 16) - (- 2) = (- 18)

459 - 18 = 441

So there are (457 - 18 =) 441  numbers that at least have one 7 or one 9 as digits.

Correct me if I am wrong. (probably wrong)

Feb 2, 2022
edited by Straight  Feb 2, 2022
edited by Straight  Feb 2, 2022
edited by Straight  Feb 2, 2022
edited by Straight  Feb 2, 2022
edited by Straight  Feb 2, 2022
#2
+676
0

ignore my message, i'll do it again

Feb 2, 2022
#3
+1315
+5

Thanks anyway:

I got 452 :)

Feb 2, 2022
#4
+676
+2

Can you show us how..?

I understand my mistake now 179, 197,,,

Feb 2, 2022
edited by Straight  Feb 2, 2022