document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1021867>Question 1021867</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I have total 3 things and 5 slots . How can i fill this 5 empty slots using only 3 things ? please answer me . thanks in advance     </SPAN>\n" );
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document.write( "Question:<BR>\n" );
document.write( "I have total 3 things and 5 slots . How can i fill this 5 empty slots using only 3 things ? please answer me . thanks in advance    <BR>\n" );
document.write( " <BR>\n" );
document.write( "Solution:<BR>\n" );
document.write( "It is obvious that there will have to be at least two empty slots, by the pigeon hole principle.<BR>\n" );
document.write( " <BR>\n" );
document.write( "The question is probably how many ways I can fill three different objects into 5 empty slots, where order of the slots is important (permutation), but the order of putting objects in the slots is not important (combinations).<BR>\n" );
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document.write( "Note that there is no limitation on how many objects can be put in one slot.<BR>\n" );
document.write( "We further assume that all three objects must end up in one slot or another (i.e. all put in the slots).<BR>\n" );
document.write( " <BR>\n" );
document.write( "So putting three objects in the same slot, there are 5 ways.<BR>\n" );
document.write( "|ABC| | | | |<BR>\n" );
document.write( "| |ABC| | | |<BR>\n" );
document.write( "... etc<BR>\n" );
document.write( "| | | | |ABC| <BR>\n" );
document.write( "5 ways..................(1)<BR>\n" );
document.write( " <BR>\n" );
document.write( "Putting two objects in one slot and one in one of the remaining slots:<BR>\n" );
document.write( "|AB|C| | | |<BR>\n" );
document.write( "|AB| |C| | |<BR>\n" );
document.write( "...<BR>\n" );
document.write( "|AB| | | |C|<BR>\n" );
document.write( "that makes 4 ways.<BR>\n" );
document.write( "However, there are 4 ways for each position of AB, from 1 to 5.  That makes 4*5=20 ways.<BR>\n" );
document.write( "In addition, instead of AB, it could have been AC or BC, so that makes 4*5*3=60 ways..........(2)<BR>\n" );
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document.write( "Now if we put the objects into individual slots, the first object has 5 choices, second 4 and third 3 for a total of 60 ways.............(3)<BR>\n" );
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document.write( "Adding up (1), (2) and (3) we have 5+60+60=125 ways.<BR>\n" );
document.write( " <BR>\n" );
document.write( "Additional Question:<BR>\n" );
document.write( "here is the question : (Q) How many ways can we arrange the letters in the word PROVIDE so that no two vowels are adjacent?  please give me the solution with explanation . thanks<BR>\n" );
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document.write( "Solution:<BR>\n" );
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document.write( "To solve this problem, we understand that<BR>\n" );
document.write( "1. When there are no restrictions, the number of arrangements is 7!=5040.<BR>\n" );
document.write( "This number includes cases where two of the vowels are together, or three vowels are together.<BR>\n" );
document.write( "2. When there are at least two vowels together, we can treat this as a word of 6 letters, giving 6!=720 arrangements.  However, since there are 6 ways of choosing the two letters (order counts), there are 6*6!=4320 cases where at least two vowels are together.  NOTE that this case (2) includes cases where all three vowels are together!<BR>\n" );
document.write( "3. When all three vowels are together, we can treat the case as a word of 5 letters, with 5!=120 arrangements.  However, there are 3!=6 ways (order counts) of choosing the 3 letters, so there are 3!*5!=720 arrangements.<BR>\n" );
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document.write( "To sum up, if we don't want any of the vowels together, we have <BR>\n" );
document.write( "# of arrangements <BR>\n" );
document.write( "= case (1)-( (2)-(3) ) <BR>\n" );
document.write( "= case (1) -(2) + (3)<BR>\n" );
document.write( "= 5040 - 4320 + 720<BR>\n" );
document.write( "= 1440<BR>\n" );
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document.write( "Looking it another way:<BR>\n" );
document.write( "The four consonants can be arranged in 4!=24 ways.<BR>\n" );
document.write( "The vowels can be \"inserted\" around the consonants in 5C3=10 ways, since the order of the vowels counts, we multiply by 3!=6.  The total number of ways is therefore 24*10*6=1440 ways as before.<BR>\n" );
document.write( "|C|C|C|C|<BR>\n" );
document.write( "vowels can be inserted in any of the five \"|\" positions, but not two in the same slot, so 5C3=5!/(3!2!)=10.<BR>\n" );
document.write( "Since ordering of vowels is important, we have {EIO,EOI,IEO,IOE,OEI,OIE}<BR>\n" );
document.write( "for a total of 6 ways = 3!.<BR>\n" );
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document.write( "In counting problems, it is always advisable to attempt counting in different ways to verify the answers against each other.<BR>\n" );
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