2

The number of $4$ digit numbers which are not divisible by $5$ that can be formed using the digits $(0,2,4,5)$ if digits are not repeated is?

Cookie
  • 13,532
Vikas
  • 21
  • The last digit can take any one of $2$ values. For each choice, the first digit can take on any one of $2$ values. And the remaining two digits can be placed in the remaining two slots in $2$ ways, for a total of $(2)(2)(2)$. – André Nicolas Jul 19 '14 at 05:43
  • 1
    What have you try? Specifically, have you considering listing out all the possibilities (there are not many), to see if you can discover any pattern? – Gina Jul 19 '14 at 05:46

3 Answers3

3

There are 12 such numbers

$0254$$\hspace{0.9cm}$ $0524$$\hspace{0.9cm}$ $2054$$\hspace{0.9cm}$ $2504$$\hspace{0.9cm}$

$5024$$\hspace{0.9cm}$ $5204$$\hspace{0.9cm}$ $0452$$\hspace{0.9cm}$ $0542$

$4052$$\hspace{0.9cm}$ $4502$$\hspace{0.9cm}$ $5042$$\hspace{0.9cm}$ $5402$

4 are 3 digit numbers$\hspace{0.9cm}$ $0254$$\hspace{0.9cm}$$0524$$\hspace{0.9cm}$$0452$$\hspace{0.9cm}$$0542$

So the answer is 8

Nannes
  • 1,141
1

Well:

  • It cannot start with $0$ (because it must be a $4$-digit number)
  • It cannot end with $0$ or $5$ (because it must not be divisible by $5$)

This leaves you with only $8$ options:

  • $2054$
  • $2504$
  • $4052$
  • $4502$
  • $5024$
  • $5042$
  • $5204$
  • $5402$
barak manos
  • 43,109
0

$$4!-(3!+3!+3!)+2!=8$$ $4!$ for all permutations,

$3!$ for the numbers starting with $0$, ending on $0$ and ending on $5$ respectively,

$2!$ for the numbers starting with $0$ and ending on $5$.

drhab
  • 151,093