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Question-120431




Question Number 120431 by john santu last updated on 31/Oct/20
Commented by john santu last updated on 31/Oct/20
old unswered question
$${old}\:{unswered}\:{question} \\ $$
Commented by mr W last updated on 31/Oct/20
question is not quite clear for me.  according to my understanding a  memorable number is e.g.  123−1234 or 123−4123.  so my answer is  2×C_3 ^(10) ×3!×7=10080
$${question}\:{is}\:{not}\:{quite}\:{clear}\:{for}\:{me}. \\ $$$${according}\:{to}\:{my}\:{understanding}\:{a} \\ $$$${memorable}\:{number}\:{is}\:{e}.{g}. \\ $$$$\mathrm{123}−\mathrm{1234}\:{or}\:\mathrm{123}−\mathrm{4123}. \\ $$$${so}\:{my}\:{answer}\:{is} \\ $$$$\mathrm{2}×{C}_{\mathrm{3}} ^{\mathrm{10}} ×\mathrm{3}!×\mathrm{7}=\mathrm{10080} \\ $$
Answered by john santu last updated on 31/Oct/20
Let A denote the set of telephone numbers  for which d_1 d_2 d_3  is the same as d_4 d_5 d_6   and let B the set of telephone numbers   for which d_1 d_2 d_3  concides with d_5 d_6 d_7 .  A telephone number d_1 d_2 d_3 −d_4 d_5 d_6 d_7   belong to A∩B if only if d_1 =d_2 =d_3 =d_4 =...=d_7 .  Hence n(A∩B)=10. Thus by Inclusion−Exclusion Principle  n(A∪B)=n(A)+n(B)−n(A∩B)                      = 10^3 .1.10 + 10^3 .10.1−10                      = 20,000−10=19,990
$${Let}\:{A}\:{denote}\:{the}\:{set}\:{of}\:{telephone}\:{numbers} \\ $$$${for}\:{which}\:{d}_{\mathrm{1}} {d}_{\mathrm{2}} {d}_{\mathrm{3}} \:{is}\:{the}\:{same}\:{as}\:{d}_{\mathrm{4}} {d}_{\mathrm{5}} {d}_{\mathrm{6}} \\ $$$${and}\:{let}\:{B}\:{the}\:{set}\:{of}\:{telephone}\:{numbers}\: \\ $$$${for}\:{which}\:{d}_{\mathrm{1}} {d}_{\mathrm{2}} {d}_{\mathrm{3}} \:{concides}\:{with}\:{d}_{\mathrm{5}} {d}_{\mathrm{6}} {d}_{\mathrm{7}} . \\ $$$${A}\:{telephone}\:{number}\:{d}_{\mathrm{1}} {d}_{\mathrm{2}} {d}_{\mathrm{3}} −{d}_{\mathrm{4}} {d}_{\mathrm{5}} {d}_{\mathrm{6}} {d}_{\mathrm{7}} \\ $$$${belong}\:{to}\:{A}\cap{B}\:{if}\:{only}\:{if}\:{d}_{\mathrm{1}} ={d}_{\mathrm{2}} ={d}_{\mathrm{3}} ={d}_{\mathrm{4}} =…={d}_{\mathrm{7}} . \\ $$$${Hence}\:{n}\left({A}\cap{B}\right)=\mathrm{10}.\:{Thus}\:{by}\:{Inclusion}−{Exclusion}\:{Principle} \\ $$$${n}\left({A}\cup{B}\right)={n}\left({A}\right)+{n}\left({B}\right)−{n}\left({A}\cap{B}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{10}^{\mathrm{3}} .\mathrm{1}.\mathrm{10}\:+\:\mathrm{10}^{\mathrm{3}} .\mathrm{10}.\mathrm{1}−\mathrm{10} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{20},\mathrm{000}−\mathrm{10}=\mathrm{19},\mathrm{990} \\ $$

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