b) very first digit - 10 ; 2nd digit - 10 ; third digit - 9 ; 4th digit - 8 --- So: $10*10*9*8= 7200$

c) very first digit - 7 ; 2nd digit - 6 ; 3rd digit - 5 ; fourth digit - 4--- So: $7*6*5*4 = 840$

Please exactly me if I"m wrong!!!


If ns am not obtaining your question wrong then your answers room wrong.a)If a 4 number number includes exactly 4 9"s , climate $^9C_4$ is not your equipment , over there is only one method i.e 9999 , so answer need to be 1 b)This one is no as basic as friend did,exactly two same will have actually conditionseither $1^st$ and $2^nd$ are same or $1^st$ and $3^rd$ are same and also so ~ above , there will certainly be 6 such conditions , girlfriend will have to solve them separately , .c)It"s nowhere pointed out that digits various other than last one can not be even so answer must be 10 x 10 x 10 x 3 , because that (4,6,8 viz. Even digits better than 2)


a) there is only one string of length $4$ that has four nines, "9999", so the prize is $1$.

You are watching: How many strings of four decimal digits end with an even digit

b) If exactly two digits room the same.: an initial choose the two places out that $4$ that space going to contain the exact same digit in $\binom42=6$ ways.

Then choose the digit that goes in those $2$ areas : $10$ choices, climate we have $9$ and $8$ options for the two continuing to be positions (in left to right order, because that definiteness). So finally we have $6 \times 10 \times 9\times 8$ methods to choose, which offers us $4320$ strings.

c) All an initial three digits are complimentary to choose, independently, the last one is one of $\4,6,8\$ so has only $3$ options. For this reason $3 \times 10^3 = 3000$ strings.


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