4
$\begingroup$

I'm having troubles with this one:

How many $5$-digit numbers can be made from digits in the number $75226522$ ?

So there is: one seven, one six, two fives, and four twos in that number.

So I guess I need to use combination with repetition but I'm having trouble with understanding this.

I know how many $8$-digit numbers there are: $$ \frac{8!}{4!2!}$$ But I'm clueless in $5$-digit case

$\endgroup$
4
  • $\begingroup$ Hint: Looks like you need to choose any 5 digits from that number, and each time you take one of the digits you are not replacing it. $\endgroup$ Aug 9, 2016 at 10:35
  • $\begingroup$ I would start off with numbers with 4 2's in them, then numbers with 3 2's (can also have 2 5's is a special case), then numbers with 2 x 2's, then 1 , then zero $\endgroup$
    – Cato
    Aug 9, 2016 at 10:42
  • $\begingroup$ 4 2's is 3 choices of 5th number, 5 ways of arranging them, 3 x 5 = 15 $\endgroup$
    – Cato
    Aug 9, 2016 at 10:48
  • $\begingroup$ 3 2's + 2 5's = 5!/2!/3! $\endgroup$
    – Cato
    Aug 9, 2016 at 10:52

2 Answers 2

2
$\begingroup$

Add up the following:

  • The number of ways to rearrange $\color\red{2222}\color\green{ 5}\color\orange{ }\color\lightblue{ }$, which is $\frac{5!}{\color\red{4!}\color\green{1!}\color\orange{ }\color\lightblue{ }}= 5$
  • The number of ways to rearrange $\color\red{2222}\color\green{ }\color\orange{6}\color\lightblue{ }$, which is $\frac{5!}{\color\red{4!}\color\green{ }\color\orange{1!}\color\lightblue{ }}= 5$
  • The number of ways to rearrange $\color\red{2222}\color\green{ }\color\orange{ }\color\lightblue{7}$, which is $\frac{5!}{\color\red{4!}\color\green{ }\color\orange{ }\color\lightblue{1!}}= 5$
  • The number of ways to rearrange $\color\red{ 222}\color\green{55}\color\orange{ }\color\lightblue{ }$, which is $\frac{5!}{\color\red{3!}\color\green{2!}\color\orange{ }\color\lightblue{ }}=10$
  • The number of ways to rearrange $\color\red{ 222}\color\green{ 5}\color\orange{6}\color\lightblue{ }$, which is $\frac{5!}{\color\red{3!}\color\green{1!}\color\orange{1!}\color\lightblue{ }}=20$
  • The number of ways to rearrange $\color\red{ 222}\color\green{ 5}\color\orange{ }\color\lightblue{7}$, which is $\frac{5!}{\color\red{3!}\color\green{1!}\color\orange{ }\color\lightblue{1!}}=20$
  • The number of ways to rearrange $\color\red{ 222}\color\green{ }\color\orange{6}\color\lightblue{7}$, which is $\frac{5!}{\color\red{3!}\color\green{ }\color\orange{1!}\color\lightblue{1!}}=20$
  • The number of ways to rearrange $\color\red{ 22}\color\green{55}\color\orange{6}\color\lightblue{ }$, which is $\frac{5!}{\color\red{2!}\color\green{2!}\color\orange{1!}\color\lightblue{ }}=30$
  • The number of ways to rearrange $\color\red{ 22}\color\green{55}\color\orange{ }\color\lightblue{7}$, which is $\frac{5!}{\color\red{2!}\color\green{2!}\color\orange{ }\color\lightblue{1!}}=30$
  • The number of ways to rearrange $\color\red{ 22}\color\green{ 5}\color\orange{6}\color\lightblue{7}$, which is $\frac{5!}{\color\red{2!}\color\green{1!}\color\orange{1!}\color\lightblue{1!}}=60$
  • The number of ways to rearrange $\color\red{ 2}\color\green{55}\color\orange{6}\color\lightblue{7}$, which is $\frac{5!}{\color\red{1!}\color\green{2!}\color\orange{1!}\color\lightblue{1!}}=60$

Hence the total number of ways is $5+5+5+10+20+20+20+30+30+60+60=265$.

$\endgroup$
1
$\begingroup$

Number with no 5's, so either three or four 2's: $$\frac{5!}{3!} + 2 \times \frac{5!}{4!}.$$ Number with one 5, so either two, three, or four 2's: $$\frac{5!}{2!} + 2 \times \frac{5!}{3!} + \frac{5!}{4!}.$$ Number with two 5's, so either one, two, or three 2's: $$\frac{5!}{2!} + 2 \times \frac{5!}{2! 2!} + \frac{5!}{2! 3!}.$$ Now adding these three numbers together should give the right result.

$\endgroup$
9
  • $\begingroup$ For the situation where there are no 5s and three 2s, I think it should be 4!/2! ? $\endgroup$
    – Mufei Li
    Aug 9, 2016 at 10:57
  • $\begingroup$ Why so? We're still permuting 5 digits... $\endgroup$
    – Mr. Chip
    Aug 9, 2016 at 10:58
  • $\begingroup$ Is there a generic solution for this type of problem. Or it is always about considering different cases ? $\endgroup$
    – tomtom
    Aug 9, 2016 at 11:00
  • $\begingroup$ Not one I'm aware of. It would certainly become painful with too many more repeated digits, but you can at least be systematic about breaking into cases. $\endgroup$
    – Mr. Chip
    Aug 9, 2016 at 11:01
  • $\begingroup$ @Mr.Chip Never mind, I made a mistake. $\endgroup$
    – Mufei Li
    Aug 9, 2016 at 11:02

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .