Question

Just another cute problem,

How many eight digit numbers formed using the digit \( \{7,8,9\} \) has exactly three \( 7 \)'s ?

Answer 1

5P2*3P1?

Answer 2

Well, Ishaan it's one of my cute problem so may be not that easy ;-)

Answer 3

Lol, Okay. I will try again.

Answer 4

sure :D

Answer 5

Btw Ishaan, I think this problem is IIT JEE standard, and might be extended to any undergrad level entrance test, so good luck! :) You can do it.

Answer 6

\[--------\]
8 slots are to be filled by three numbers {7,8,9}, we need numbers who have exactly 3 7s. Implies, 5 slots left for {8,9}, order matters, 5P2 must be the number of ways slots can be filled. 3 slots left, and I am supposed to fill them up with 7s, order doesn't matter, so only 1 way.
5P2, is the answer 20?

Answer 7

I did something wrong, I need to get hold on Combinatorics.

Answer 8

Oh wait, the choice of filling 5slots with numbers 8 and 9, depends upon the arrangement of 3 slots (which are to be filled by 7s). So, 8C3*5P2.

I am not sure again, just trying to improvise.

Answer 9

\[2^{5}\cdot{8\choose 3}\]

Answer 10

That's gives the correct answer @Zarkon

Actually my approach is a bit mildly tedious so you may like to explain your approach for us mortal.

Answer 11

if there are 8 digits and 3 of them must be 7's then 5 must be 8's or 9's

so for each of those 5 numbers there are 2 choices. thus \(2^5\) options.

now we just choose the 3 places to put the sevens. out of the 8 possible position we choose 3.

Answer 12

But 2^{5} would include repetition of the numbers, wouldn't it?

Answer 13

no...there are only 32 (not including the 7's) of them...list them out to convince yourself.

Answer 14

Oh, thanks zarkon.

Answer 15

My solution is : \[ \Huge \sum \limits_{n= 0}^5 \frac{8!}{3! \times n! \times {(5-n)!}} \]

Answer 16

\[ \sum \limits_{n= 0}^5 \frac{8!}{3! \times n! \times {(5-n)!}}\]
\[=\sum \limits_{n= 0}^5 \frac{8!\times 5!}{3!\times 5! \times n! \times {(5-n)!}}\]
\[=\frac{8!}{3!\times 5!}\sum \limits_{n= 0}^5 \frac{5!}{n! \times {(5-n)!}}\]
\[={8\choose 3}\sum \limits_{n= 0}^5 \frac{5!}{n! \times {(5-n)!}}\]
\[={8\choose 3}\sum \limits_{n= 0}^5 \frac{5!}{n! \times {(5-n)!}}1^n1^{5-n}\]
\[={8\choose 3}(1+1)^5\]
\[={8\choose 3}\cdot2^5\]

Answer 17

@Zarkon: Great answer, I like the way you reduced \( \large \sum \limits_{n= 0}^5 \frac{5!}{n! \times {(5-n)!}} \; to \; 2^5 \) using binomial theorem!

Answer 18

yep..cute problem.

Answer 19

Something happened to latex, is it rendering on server side now?

Answer 20

Cutie pie :D

btw we could also do this \[ \large \sum \limits_{n= 0}^5 \frac{5!}{n! \times {(5-n)!}} = \large \sum \limits_{n= 0}^5\binom{5}{n} = 2^5\]

Using the identity, \(\large \sum \limits_{r= 0}^n\binom{n}{r} = 2^n \)

Answer 21

Just to triger the notification system @Zarkon :-)

Answer 22

@Ishaan94: No problem in my side.

Answer 23

which again...is really the binomial theorem

Answer 24

yep!

Answer 25

We just need to expand \( (1+x)^n\) using binomial theorem and substitute \(x=1\) that will gives us the desired identity.