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Question Number 39846 by vajpaithegrate@gmail.com last updated on 12/Jul/18
the sum of the four  digit even numbers that can be formed with digits 0 3 5 4 without repitation
$$\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{four}\:\:\mathrm{digit}\:\mathrm{even}\:\mathrm{numbers}\:\mathrm{that}\:\mathrm{can}\:\mathrm{be}\:\mathrm{formed}\:\mathrm{with}\:\mathrm{digits}\:\mathrm{0}\:\mathrm{3}\:\mathrm{5}\:\mathrm{4}\:\mathrm{without}\:\mathrm{repitation} \\ $$
Commented by Rasheed.Sindhi last updated on 13/Jul/18
43536
$$\mathrm{43536} \\ $$
Commented by vajpaithegrate@gmail.com last updated on 13/Jul/18
how sir pls explain
$$\mathrm{how}\:\mathrm{sir}\:\mathrm{pls}\:\mathrm{explain} \\ $$
Answered by Rasheed.Sindhi last updated on 13/Jul/18
∵ The numbers are even  ∴ The unit place can be filled with 0 or 4  Case: − − − 0  When unit is 0  tens can be filled in 3 ways (3,4 or 5)  after that hundreds can be filled in 2 ways  and thousands can be filled in 1 way only.  ∴Total numbers with unit 0:        1×3×2×1=6  Case: − − − 4  When unit is 4  thousands can be filled in 2 ways(3 or 5)  (It can′t be 0 as in that case the numbers  will be of 3 digits)  hundres can be filled in 2 ways and   tens can be filled in 1 way.  ∴ Total numbers with unit 4:      1×2×2×1=4  So sum of units of all the 10 numbers is  0×6+4×4=16 (6 units & 1 carry to tens.  Case: − − 0 −  When tens is 0→unit can be filled in 1 way(with4)  →hundreds can be filled in 2 ways (3 or 5)  →thousands can be filled in 1 way.  ∴Total numbers with tens 0        1×1×2×1=2  Case:− − 4 −       1×1×2×1=2  Case: − − 3 −      2×1×  Continue
$$\because\:\mathrm{The}\:\mathrm{numbers}\:\mathrm{are}\:\mathrm{even} \\ $$$$\therefore\:\mathrm{The}\:\mathrm{unit}\:\mathrm{place}\:\mathrm{can}\:\mathrm{be}\:\mathrm{filled}\:\mathrm{with}\:\mathrm{0}\:\mathrm{or}\:\mathrm{4} \\ $$$$\mathrm{Case}:\:−\:−\:−\:\mathrm{0} \\ $$$$\mathrm{When}\:\mathrm{unit}\:\mathrm{is}\:\mathrm{0} \\ $$$$\mathrm{tens}\:\mathrm{can}\:\mathrm{be}\:\mathrm{filled}\:\mathrm{in}\:\mathrm{3}\:\mathrm{ways}\:\left(\mathrm{3},\mathrm{4}\:\mathrm{or}\:\mathrm{5}\right) \\ $$$$\mathrm{after}\:\mathrm{that}\:\mathrm{hundreds}\:\mathrm{can}\:\mathrm{be}\:\mathrm{filled}\:\mathrm{in}\:\mathrm{2}\:\mathrm{ways} \\ $$$$\mathrm{and}\:\mathrm{thousands}\:\mathrm{can}\:\mathrm{be}\:\mathrm{filled}\:\mathrm{in}\:\mathrm{1}\:\mathrm{way}\:\mathrm{only}. \\ $$$$\therefore\mathrm{Total}\:\mathrm{numbers}\:\mathrm{with}\:\mathrm{unit}\:\mathrm{0}: \\ $$$$\:\:\:\:\:\:\mathrm{1}×\mathrm{3}×\mathrm{2}×\mathrm{1}=\mathrm{6} \\ $$$$\mathrm{Case}:\:−\:−\:−\:\mathrm{4} \\ $$$$\mathrm{When}\:\mathrm{unit}\:\mathrm{is}\:\mathrm{4} \\ $$$$\mathrm{thousands}\:\mathrm{can}\:\mathrm{be}\:\mathrm{filled}\:\mathrm{in}\:\mathrm{2}\:\mathrm{ways}\left(\mathrm{3}\:\mathrm{or}\:\mathrm{5}\right) \\ $$$$\left(\mathrm{It}\:\mathrm{can}'\mathrm{t}\:\mathrm{be}\:\mathrm{0}\:\mathrm{as}\:\mathrm{in}\:\mathrm{that}\:\mathrm{case}\:\mathrm{the}\:\mathrm{numbers}\right. \\ $$$$\left.\mathrm{will}\:\mathrm{be}\:\mathrm{of}\:\mathrm{3}\:\mathrm{digits}\right) \\ $$$$\mathrm{hundres}\:\mathrm{can}\:\mathrm{be}\:\mathrm{filled}\:\mathrm{in}\:\mathrm{2}\:\mathrm{ways}\:\mathrm{and}\: \\ $$$$\mathrm{tens}\:\mathrm{can}\:\mathrm{be}\:\mathrm{filled}\:\mathrm{in}\:\mathrm{1}\:\mathrm{way}. \\ $$$$\therefore\:\mathrm{Total}\:\mathrm{numbers}\:\mathrm{with}\:\mathrm{unit}\:\mathrm{4}: \\ $$$$\:\:\:\:\mathrm{1}×\mathrm{2}×\mathrm{2}×\mathrm{1}=\mathrm{4} \\ $$$$\mathrm{So}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{units}\:\mathrm{of}\:\mathrm{all}\:\mathrm{the}\:\mathrm{10}\:\mathrm{numbers}\:\mathrm{is} \\ $$$$\mathrm{0}×\mathrm{6}+\mathrm{4}×\mathrm{4}=\mathrm{16}\:\left(\mathrm{6}\:\mathrm{units}\:\&\:\mathrm{1}\:\mathrm{carry}\:\mathrm{to}\:\mathrm{tens}.\right. \\ $$$$\mathrm{Case}:\:−\:−\:\mathrm{0}\:− \\ $$$$\mathrm{When}\:\mathrm{tens}\:\mathrm{is}\:\mathrm{0}\rightarrow\mathrm{unit}\:\mathrm{can}\:\mathrm{be}\:\mathrm{filled}\:\mathrm{in}\:\mathrm{1}\:\mathrm{way}\left(\mathrm{with4}\right) \\ $$$$\rightarrow\mathrm{hundreds}\:\mathrm{can}\:\mathrm{be}\:\mathrm{filled}\:\mathrm{in}\:\mathrm{2}\:\mathrm{ways}\:\left(\mathrm{3}\:\mathrm{or}\:\mathrm{5}\right) \\ $$$$\rightarrow\mathrm{thousands}\:\mathrm{can}\:\mathrm{be}\:\mathrm{filled}\:\mathrm{in}\:\mathrm{1}\:\mathrm{way}. \\ $$$$\therefore\mathrm{Total}\:\mathrm{numbers}\:\mathrm{with}\:\mathrm{tens}\:\mathrm{0} \\ $$$$\:\:\:\:\:\:\mathrm{1}×\mathrm{1}×\mathrm{2}×\mathrm{1}=\mathrm{2} \\ $$$$\mathrm{Case}:−\:−\:\mathrm{4}\:− \\ $$$$\:\:\:\:\:\mathrm{1}×\mathrm{1}×\mathrm{2}×\mathrm{1}=\mathrm{2} \\ $$$$\mathrm{Case}:\:−\:−\:\mathrm{3}\:− \\ $$$$\:\:\:\:\mathrm{2}×\mathrm{1}× \\ $$$$\mathrm{Continue} \\ $$
Answered by Rasheed.Sindhi last updated on 13/Jul/18
  5430 + 4530 + 5340 + 3540 + 4350  + 3450 + 5304 + 3504 + 5034 +3054  = 43536  Note: The numbers have been achieved  by tree-diagram,which can′t be drawn  by forum-keyboard.
$$\:\:\mathrm{5430}\:+\:\mathrm{4530}\:+\:\mathrm{5340}\:+\:\mathrm{3540}\:+\:\mathrm{4350} \\ $$$$+\:\mathrm{3450}\:+\:\mathrm{5304}\:+\:\mathrm{3504}\:+\:\mathrm{5034}\:+\mathrm{3054} \\ $$$$=\:\mathrm{43536} \\ $$$$\mathrm{Note}:\:\mathrm{The}\:\mathrm{numbers}\:\mathrm{have}\:\mathrm{been}\:\mathrm{achieved} \\ $$$$\mathrm{by}\:\mathrm{tree}-\mathrm{diagram},\mathrm{which}\:\mathrm{can}'\mathrm{t}\:\mathrm{be}\:\mathrm{drawn} \\ $$$$\mathrm{by}\:\mathrm{forum}-\mathrm{keyboard}. \\ $$

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