Question

Se considera multimea M={1,2,3,....,11}.Determinati numărul submultimilor de forma {a, b, c, d}, ale multimii M, cu proprietatea a+b=c+d.
Va rog ajutati-ma!! Dau coroana!! ​​

Answer 1

Răspuns:

Explicație pas cu pas:

ttrebuiesa gasim multimi de 4 elente compuse din 2 perechi care sa indeplinasca cerinta

suma a doua elete poate fi:

suma min=3=1+2=3 nu cinvine

suma=1+3=4 nu convine, nu exista alta combinatiede 2 elem ente diferite

suma=1+4=5=2+3 un cvartete fen {1.2;3;4} Comb de 2 luate cate 2=1 cvartet

suma=1+5=6=2+4 Comb de 2 luate cate 2=1 cvartet {1;2;4;5}

suma=7=1+6=3+4=2+5 Combinari de 3 liuatecate2= 3 cvartete

suma=1+7=8=2+6=3+5...idem, Comb de 3 luate cate 2=3 cvartete

suma=9=1`+8=2+7=3+6=4+5 .idem, Comb de 4 luate cate 2=6 cvartete.

suma =10=1+9=2+8=3+7=4+6 Comb de 4 luate cate 2=6 cvartete

suma=11=1+10=2+9=3+8=4+7=5+6 Comb de 5 lauatecate 2=15 cvartete

suma=12=1`+11=2+10=3+9=4+8=5+7 Comb de 5 lauatecate 2=15 cvartete

suma=13=2+11=3+10=4+9=5+8=6+7 Comb de 5 lauatecate 2=15 cvartete

suma=14=3+11=4+10=5+9=6+8..Comb de 4 luate cate 2=6 cvartete

suma=15=4+11=5+10=6+9=7+8...Comb de 4 luate cate 2=6 cvartete

suma=16=5+11=6+10=7+9...Combinari de 3 liuatecate2= 3 cvartete

suma=17=6+11=7+10=8+9 .Combinari de 3 liuatecate2= 3 cvartete

suma=18=7+11=8+10= Comb de 2 luate cate 2=1 cvartet

suma=19=11+8 =10+9 ..Comb de 2 luate cate 2=1 cvartet

probabil e bine, distributia estes imeterica

suam=20=11+9 nu sev poate gasi alta pecereche de nunmere in jmultimea data

Total...le aduni si tu,...partea grea e facuta