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POhow to get number of possible 4 digit numbers with constraints on possible digits and position of digits
 

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    <p>I was trying to solve a programming problem and got stuck because I couldn't understand one of the examples which goes as follows, </p> <p>We guess a four digit number and the guess is "1234" . The hints given for this guess are, </p> <ol> <li><p>Each of the digits is not in it's right place (as per correct answer). That is , 1 is not in position 1, 2 is not in position 2, 3 is not in position 3 and 4 is not in position 4. </p></li> <li><p>The 4 digit correct answer contains the digits 1,2,3,4. </p></li> </ol> <p>The example gave the number of possible combinations of four digit numbers based on above constraints to be 9. {2143,2341,2413,3142,3412,3421,4123,4312,4321}</p> <p>I tried to approach the problem this way:</p> <p>approach1 :</p> <p>(Total # of combinations which is 4!) - (( Combinations which start with 1 in position 1 + Combinations with 2 in position 2 + Combinations with 3 in position 3 + Combinations with 4 in position 4)) but Couldn't get to a solution for second part of the above formula.. as Combinations which start with 1 in position 1 will be 3!- (combination which start with 2 in position 2).. and so on and I wasn't able to proceed on writing the number of combinations).</p> <p>approach2:</p> <p>(1 can be in 3 positions) * (2 can be in 3 or 2 positions based on where 1 is placed) * (3 can be in 1 or 2 positions based on where 2 is place)*(1 position for 4) -- Again not clear on how to find the # of positions for 2,3,4.</p> <p>Please help me understand how to approach this problem</p>
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    1. COIt's a permutation not a combination. so the number of possible arrangments is 4! = 4 * 3 * 2 * 1 = 24
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    2. COThe possible arrangements (4!) will include ones which voilate hint 1 . So they have to be eliminated from the 4!. Trying to find a way to formulate elimination arrangements...
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    3. COI just understand it now
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