How many positive integer n can we form using the digits 2 - 2 - 2 - 5 - 6 - 6 - 7-9 ?and number given cannot?

Question

and number given cannot exceed 65,500,000

8 digit NUMBER

Answer 1

There are 8!/((3!)(2!)) possible combinations, = 3360.

Exactly 1/4 of those start off with 7 or 9, so we subtract (1/4)(3360)=840. We are left with 2520 choices.

Exactly 1/4 of the numbers start with 6. Out of these, (3/7) have an invalid next digit (6,7,9). So we subtract by (1/4)(3/7)(3360) = 360. We are left with 2160 choices.

Exactly (1/4)(1/7) of the numbers start with 65. Out of these (1/2) have an invalid next digit (6,7,9). So we subtract (1/4)(1/7)(1/2)(3360) = 60 . This leaves us with exactly 2100 combinations.

Answer 2

geeze - wish i could remember what I learned in Statistics
You've got 8 digits, 5 different numbers - um, 2, 5, 6, 7, 9, 25, 26, 27, 29, 52, 56, 22, 57, 59, 62, 65, 66, 67, 69, 72, 75, 76, 79, 92, 95, 96, 97, 222, 225, 226, 227, 229, 252, 256, 257, 259, 262, 267, 269, 272, 275, 276, 279, 292, 295, 297, 522, 526, 527, 529, 52626279, 2567292, 9722, 957266, you're on you own buddy

good luck!!!!

Answer 3

WE HAVE TO MAKE CASES
1. numbers starting with 2 : 7!/(2!2!)
2. no.s starting with 5 : 7!(3!2!)
3. no.s starting with 6:
3.A. no.s starting with 62 : 6!/2!
3.B. no.s starting with 65 :
3.B.a no.starting with 652 : 5!/2!
so total is 1 + 2 + 3.A +3.B.a =2100

Answer 4

It is sigma i=1to7 NPi + N factorial numbers. where N is the number of digits, P is the permutaiton, the last series has n factorial combinations.