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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!!!

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2 Answers 2

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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)

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  • 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, 2019 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, 2019 at 10:28
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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.

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  • $\begingroup$ @N.F.Taussig you could have edited it too, I don’t mind. Thx. $\endgroup$ Feb 3, 2019 at 10:31
  • $\begingroup$ Thank you so much.. $\endgroup$
    – Layla
    Feb 3, 2019 at 10:32

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