Question

Determine how many numbers greater than 10 can be made using all of the digits 4, 7, 2, 6 and 5, if each digit cannot be used more than once.

Answer 1

Choose 5 + choose 4 + choose 3 etc.

Answer 2

is it (20 + 60 +120+120)=320

Answer 3

u can make 2-digits , 3-digits ,...,5-digits numbers
for example how many 2-digit numbers can be made using {2,4,5,6,7} with considering that none of them can not be used more than once:
well choose 2 of {2,4,5,6,7} there is C(2,5) way for do that ; and now there is 2! permutations of 2 number choosen; so 2-digits number can be made : 2*C(2,5)

Answer 4

yes it is 320, but can u explain ur working?

Answer 5

actually my way gives
C(2,5) * 2!+C(3,5) * 3!+C(4,5) * 4!+C(5,5) * 5!=320
try to work it out from my previous reply

Answer 6

I don't quite get ur working of "well choose 2 of {2,4,5,6,7} there is C(2,5) way for do that ; and now there is 2! permutations of 2 number choosen; so 2-digits number can be made : 2*C(2,5)"

Answer 7

ok. let me explain from scratch

Answer 8

1) 2 digit numbers :
2 places can be filled in \(\color{purple}{5*4}\) ways. (for tens placce, you got 5 choices, after choosing that, for units place you left with 4 choices, since repition is not allowed)

Answer 9

ok.

Answer 10

Order matters, perm n!/(n-k)! versus comb n!/k!(n-k)!

Answer 11

Next,
2) 3 digit numbers :
3 places can be filled in \(\color{purple}{5*4*3}\) ways. (for hundreds placce, u have 5 choices, after that, for tens place 4 choices, lastly for units place - 3 choices)

Answer 12

\[5P2+5P3+5P4+5P5=\frac{5!}{3!}+\frac{5!}{2!}+\frac{5!}{1!}+\frac{5!}{1}=320\]

Answer 13

ok

Answer 14

Next,
3) 4 digit numbers :
4 places can be filled in \(\color{purple}{5*4*3*2}\) ways.

5 digit numbers :
5 places can be filled in \(\color{purple}{5*4*3*2*1}\) ways.

Answer 15

yep, I get it now thanks.

Answer 16

I appreciate u for helping me as well as having the time and patience to do so.

Answer 17

np :)