Question

How many possible 4-digit combinations are there with the numbers 2, 3, 4, 5, 6, 7, 8, and 9 if none of the numbers appear more than once (i.E. 2343, 2333, 2323, etc.)?

Answer 1

Answer:

$1680$

Step-by-step explanation:

we need a 4-digit number from the numbers $2,3,4,5,6,7,8\ and\ 9$ (without repetition ).

possible number at thousand place $=8$

Possible numbers at hundred place$=8-1=7$

Possible numbers at $10^{th}$ place $=7-1=6$

possible number at unit place $=6-1=5$

So total possible numbers

$=8\times7\times6\times5\\=1680$

Other method :

We are taking $4$ numbers out of $8$ and here order matters so we will use permutation.

Total possible numbers $=^8P_{4}$

$\frac{8!}{(8-4)!}\\ =\frac{8!}{4!}\\ =8\times7\times6\times5\\=1680$