document.write( "Question 883536: how many 3 digit numbers divisible by 5 can be formed from 1,2,3,4,5,6 ? \n" ); document.write( "

Algebra.Com's Answer #533600 by LinnW(1048) $\"\"$ $\"About$

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Since the 3 digit number must end in 5
\n" ); document.write( "we have
\n" ); document.write( "125
\n" ); document.write( "135
\n" ); document.write( "145
\n" ); document.write( "165
\n" ); document.write( "215
\n" ); document.write( "235
\n" ); document.write( "245
\n" ); document.write( "265
\n" ); document.write( "315
\n" ); document.write( "325
\n" ); document.write( "345
\n" ); document.write( "365
\n" ); document.write( "415
\n" ); document.write( "425
\n" ); document.write( "435
\n" ); document.write( "465
\n" ); document.write( "615
\n" ); document.write( "625
\n" ); document.write( "635
\n" ); document.write( "645
\n" ); document.write( "This enumeration gives us 20 possibilities
\n" ); document.write( "We could also use the permutation
\n" ); document.write( "formula $\"+n%21%2F%28n-r%29%21\"$
\n" ); document.write( "where n is the different elements ( 1,2,3,4,6)
\n" ); document.write( "and r is the number taken or 2 elements in our case
\n" ); document.write( " $\"+n%21%2F%28n-r%29%21\"$ = $\"+5%21%2F%285-2%29%21\"$
\n" ); document.write( "= $\"+5%21%2F%283%29%21\"$
\n" ); document.write( "= (5*4*3*2*1)/(3*2*1)
\n" ); document.write( "= 5*4
\n" ); document.write( "= 20 \n" ); document.write( "

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