0
$\begingroup$

a) Exactly contains four 9-s.

b) Exactly 2 digit must be same

c) The string ends with a even digit greater than 2

My solutions:

a) $\binom9 4= \frac{9!}{4!(9-4)!}= 126 $

b) First digit - 10 ; Second digit - 10 ; Third digit - 9 ; Fourth digit - 8 --- So: $10*10*9*8= 7200$

c) First digit - 7 ; Second digit - 6 ; Third digit - 5 ; Fourth digit - 4 --- So: $7*6*5*4 = 840$

Please correct me if I'm wrong!!!

$\endgroup$
3
$\begingroup$

If I am not getting your question wrong then your answers are wrong.
a)If a 4 digit number contains exactly four 9's , then $^9C_4$ is not your solution , there is only one way i.e 9999 , so answer should be 1
b)This one is not as simple as you did,exactly two same will have conditions either $1^{st}$ and $2^{nd}$ are same or $1^{st}$ and $3^{rd}$ are same and so on , there will be 6 such conditions , you will have to solve them separately , .
c)It's nowhere mentioned that digits other than last one cannot be even so answer should be 10 x 10 x 10 x 3 , for (4,6,8 viz. even digits greater than 2)

$\endgroup$
2
  • 1
    $\begingroup$ The title refers to strings of four digits, not numbers containing four digits. The first digit can be zero in a string. $\endgroup$
    – nickgard
    Feb 3 '19 at 10:06
  • $\begingroup$ When I had answered the question , it was numbers and not decimals , the question was edited i guess. $\endgroup$ Feb 3 '19 at 10:28
1
$\begingroup$

a) There is only one string of length $4$ that contains four nines, "9999", so the answer is $1$.

b) If exactly two digits are the same.: first choose the two places out of $4$ that are going to contain the same digit in $\binom{4}{2}=6$ ways.

Then pick the digit that goes in those $2$ places : $10$ choices, then we have $9$ and $8$ choices for the two remaining positions (in left to right order, for definiteness). So finally we have $6 \times 10 \times 9\times 8$ ways to choose, which gives us $4320$ strings.

c) All first three digits are free to choose, independently, the last one is one of $\{4,6,8\}$ so has only $3$ options. So $3 \times 10^3 = 3000$ strings.

$\endgroup$
2
  • $\begingroup$ @N.F.Taussig you could have edited it too, I don’t mind. Thx. $\endgroup$ Feb 3 '19 at 10:31
  • $\begingroup$ Thank you so much.. $\endgroup$
    – Layla
    Feb 3 '19 at 10:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.