6 white and 6 black balls of same size are distributed among 10 urns so that there is atleast one ball in each urn. What are number of different distributions of balls?A) 25000. B)26250. C)28250. D)13125

• The solution in short is
  
  2*10C6 - (10C6 * 6C2) + (10C6* 10C1)*2+ (10C5*5C1)*5C1=18900-3150+4200+6300
  =26250
  
  The details are as following
  
  First of all note this very follwoing important point.
  
  Whenever I distribute the balls
  Every case I consider should be completely disjoint
  That is my cases should never oeverlap
  It will get clearer as I proceed
  
  
  Supposing I label the 10 urns as 1,2,3,4,5,6,7,8,9,10.
  Observation one : Urns are distinguishable
  
  I have 6 red and 6 black balls
  
  Observation two : 6 blacks are not distinguishable among themselves
  6 whites are not ditinguishable among themselves
  
  The aim is to distribute the balls so as each distibution which
  we count has to be different in some sense
  1) Each Urn has a different number of balls
  2) if the urn has same number of balls, for each urn in two different distributions , then colour of a ball must be different for at least one urn
  
  Say number of balls in a distribution D1 and D2 urn
  is such that 1,2,3,4,5,6,7,8,9,10 have exactly equal number of balls then
  at least one of the urns must have ball of different colour
  only then I can call D1 and D2 as different and count them.
  
  Let us proceed
  
  Supposing I distribute this way
  
  Case 1
  
  I choose ANY 6 urns then I fill them with black balls.
  I get 6 filled and 4 empty
  I fill the 4 remaining empty urns with 4 RED balls
  So I have 2 RED balls left
  I choose any 2 of the 10 urns each having a single red or black ball and place the remaining 2 red balls
  I get for this
  10C6 that is choosing 6 urns out of 10
  4 urns are always empty I need not choose them.
  But the 2 urns for remaining 2 Red Balls can be chosen in 10C2 ways
  So I get for whole process
  10C6*10C2
  Now I started out with balck balls
  What if I started out with RED balls.
  I proceed identically and choose 6 urns for RED and then distribute the rest of 2 blacks among any 2 of 10 urns.
  So I get 2* (10C6*10C2) for red and balck for 6 and 2 kind of distribution
  But wait
  Supposing I distributed 6 red in urns 1,2,3,4,5,6
  And 4 blacks in 7,8,9,10
  I have 2 blacks supposing I put them in urns 5,6
  problem is that this kind of distribution was possible from other way around as well that is
  I distribute 6 Black balls in urns 5,6,7,8,9,10
  4 reds in 1,2,3,4 and remaining 2 reds I put in 5,6
  There's a double counting I need to elimiante
  the problem only occurs if I put remaining last two balls in urns containing ball of a different colour.
  So I subtract
  10C6*6C2 to eliminate double counting
  because after I put 4 balls in urns of a particular colour the last two balls can be put in urn containing balls of different colur in 6C2 ways
  
  
  
  Case 2
  
  I put 6 red balls in any 6 urns
  then 4 black in remaining 4 empty urns
  and remining 2 black balls together in any of the 10 urns having 1 ball of any colour
  
  this can be done in 10C6*10C1 ways
  And since I can start with black instead of red I multiply by 2
  to get
  (10C6* 10C1)*2
  
  there will be no double counting you can convince yourself.
  
  
  Case 3
  
  I choose 5 red balls and put them in any urns
  Then I choose 5 black balls and put them in rest of 5 empty urns
  Then I choose 1 remaining red ball and place it in urn having red ball
  Then I choose 1 remaining black ball and place it in urn having black ball
  that is red goes to red and black goes to black
  
  5 black can be chosen 10C5 ways
  rest 5 are automatically chosen I don't need to choose again for red
  the one remaining black can be put in any of 5 urns containg black in 5C1 ways
  and 1 remaing red can be put in any urn contain red in 5C1 ways
  
  I get in total (10C5*5C1)*5C1
  
• 11 years agoHelpfull: Yes(15) No(3)
• @PAWAN could u please elaborate your answer...
• 11 years agoHelpfull: Yes(6) No(0)
• b)26250
  2*10C6 - (10C6 * 6C2) + (10C6* 10C1)*2+ (10C5*5C1)*5C1=18900-3150+4200+6300
  =26250
  
• 11 years agoHelpfull: Yes(3) No(16)
• i am unable to get it...can anyone plz explain it again
  
• 10 years agoHelpfull: Yes(2) No(3)
• b)26250
  2*(10C6*10C2) - (10C6 * 6C2) + (10C6* 10C1)*2+ (10C5*5C1*5C1)=18900-3150+4200+6300
  =26250
• 11 years agoHelpfull: Yes(1) No(1)
• Answer is 26250
  
  The following assumptions are taken for this answer
  a. All white balls are identical, all black balls are identical.
  b. All urns are distinct
  c. The order in which balls are placed inside the urn does not matter
  
  10 distinct urns are there, say 1,2,....10. Each urn must have at least one ball.
  
  Case 1 : Any of the 6 urns contain one white ball each. Rest of the 4 urns contain one black ball each. Remaining two black balls are distributed in any two urns.
  
  6 urns can be selected in 10C6 ways.
  Put one white ball in each of the selected 6 urns. Since the white balls are identical, this can be done only in 1 way
  4 urns can be selected from the remaining 4 urns in 4C4 ways .
  Put one black ball in each of the selected 4 urns. 1 way of doing this
  Select any two urns in 10C2 ways
  Place two black balls in these urns. 1 way of doing so
  
  Total number of ways = 10C6 × 4C4 × 10C2
  
  Case 2 : Any of the 6 urns contain one black ball each. Rest of the 4 urns contain one white ball each. Remaining two white balls are distributed in any two urns
  
  Similar to what we have seen in case 1, this can also be done in 10C6 × 4C4 × 10C2 ways
  
  But case 2 and case 1 together has resulted in overcounting of some of the possibilities . Suppose, in case 1, a white ball is placed in each of the 1-6 urns, a black ball is placed in each of the 7-10 urns and remaining two black ball is placed in 5 and 6, as given below
  
  1 2 3 4 5 6 7 8 9 10
  W W W W W W B B B B
  B B
  
  Assume, in case 2, a black ball is placed in each of the 5-10 urns, a white ball is placed in each of the 1-4 urns and remaining two white balls are placed in 5 and 6 urns,as given below
  
  1 2 3 4 5 6 7 8 9 10
  W W W W B B B B B B
  W W
  
  As we observe, such possibilities which are already counted in case1, have been counted in case 2 as well. This overcounting occurs in case 2, when the last two white balls are placed in urns which has a black ball. Count of such possibilities are 10C6 × 4C4 × 6C2. This need to be reduced from the total to balance the effect of overcounting.
  
  Case 3 : Any of the 6 urns contain one white ball each. Rest of the 4 urns contain one black ball each. Remaining two black balls are placed together in any one urn.
  
  6 urns can be selected in 10C6 ways.
  Put one white ball in each of the selected 6 urns. 1 way of doing this
  4 urns can be selected from the remaining 4 urns in 4C4 ways .
  Put one black ball in each of the selected 4 urns. 1 way of doing this
  Select any one urn in 10C1 ways
  Place two black balls in these urns. 1 way of doing so
  
  Total number of ways = 10C6 × 4C4 × 10C1
  
  Case 4 : Any of the 6 urns contain one black ball each. Rest of the 4 urns contain one white ball each. Remaining two white balls are placed together in any one urn
  
  As seen in case 3, this gives another 10C6 × 4C4 × 10C1 possibilities
  
  Unlike what happened in case 1 and case 2, no overcounting occurs here
  
  Case 5 : Any of the 5 urns contain one white ball each. Rest of the 5 urns contain one black ball each. Remaining one white ball is placed in any of the 5 urns which already has a white ball. Remaining one black ball is placed in any of the 5 urns which already has a black ball.
  
  (Note that if we place the last white ball and/or the last black ball in urns where a different colour ball is already placed, it will result in overcounting of what we have already taken in case 1, case 2, case 3 and case 4)
  
  5 urns can be selected in 10C5 ways.
  Put one white ball in each of the selected 5 urns. 1 way of doing this
  Remaining 5 urns can be selected in 5C5 ways.
  Put one black ball in each of these 5 urns. 1 way of doing this
  Select any one urn which has a white ball in 5C1 ways
  place the remaining white ball in this urn. 1 way of doing this
  Select any one urn which has a black ball in 5C1 ways
  place the remaining black ball in this urn. 1 way of doing this
  
  Total number of ways = 10C5 × 5C5 × 5C1 × 5C1
  
  Therefore, required number of ways
  
  = (10C6×4C4×10C2)+(10C6×4C4×10C2)-(10C6×4C4×6C2)+(10C6×4C4×10C1)
  +(10C6×4C4×10C1)+(10C5×5C5×5C1×5C1)
  
  = (10C6×10C2)+(10C6×10C2)-(10C6×6C2)+(10C6×10C1)+(10C6×10C1)+(10C5×5C1×5C1)
  
  = 2(10C6×10C2)-(10C6×6C2)+2(10C6×10C1)+(10C5×5C1×5C1)
  
  = 2(210×45)-(210×15)+2(210×10)+(252×5×5)
  
  = 18900 - 3150 + 4200 + 6300
  = 26250
• 7 years agoHelpfull: Yes(1) No(0)
• http://www.careerbless.com/qna/discuss.php?questionid=952
• 7 years agoHelpfull: Yes(0) No(0)
• Answer is 26250
  
  The following assumptions are taken for this answer
  a. All white balls are identical, all black balls are identical.
  b. All urns are distinct
  c. The order in which balls are placed inside the urn does not matter
  
  10 distinct urns are there, say 1,2,....10. Each urn must have at least one ball.
  
  Case 1 : Any of the 6 urns contain one white ball each. Rest of the 4 urns contain one black ball each. Remaining two black balls are distributed in any two urns.
  
  6 urns can be selected in 10C6 ways.
  Put one white ball in each of the selected 6 urns. Since the white balls are identical, this can be done only in 1 way
  4 urns can be selected from the remaining 4 urns in 4C4 ways .
  Put one black ball in each of the selected 4 urns. 1 way of doing this
  Select any two urns in 10C2 ways
  Place two black balls in these urns. 1 way of doing so
  
  Total number of ways = 10C6 × 4C4 × 10C2
  
  Case 2 : Any of the 6 urns contain one black ball each. Rest of the 4 urns contain one white ball each. Remaining two white balls are distributed in any two urns
  
  Similar to what we have seen in case 1, this can also be done in 10C6 × 4C4 × 10C2 ways
  
  But case 2 and case 1 together has resulted in overcounting of some of the possibilities . Suppose, in case 1, a white ball is placed in each of the 1-6 urns, a black ball is placed in each of the 7-10 urns and remaining two black ball is placed in 5 and 6, as given below
  
  1 2 3 4 5 6 7 8 9 10
  W W W W W W B B B B
  B B
  
  Assume, in case 2, a black ball is placed in each of the 5-10 urns, a white ball is placed in each of the 1-4 urns and remaining two white balls are placed in 5 and 6 urns,as given below
  
  1 2 3 4 5 6 7 8 9 10
  W W W W B B B B B B
  W W
  
  As we observe, such possibilities which are already counted in case1, have been counted in case 2 as well. This overcounting occurs in case 2, when the last two white balls are placed in urns which has a black ball. Count of such possibilities are 10C6 × 4C4 × 6C2. This need to be reduced from the total to balance the effect of overcounting.
  
  Case 3 : Any of the 6 urns contain one white ball each. Rest of the 4 urns contain one black ball each. Remaining two black balls are placed together in any one urn.
  
  6 urns can be selected in 10C6 ways.
  Put one white ball in each of the selected 6 urns. 1 way of doing this
  4 urns can be selected from the remaining 4 urns in 4C4 ways .
  Put one black ball in each of the selected 4 urns. 1 way of doing this
  Select any one urn in 10C1 ways
  Place two black balls in these urns. 1 way of doing so
  
  Total number of ways = 10C6 × 4C4 × 10C1
  
  Case 4 : Any of the 6 urns contain one black ball each. Rest of the 4 urns contain one white ball each. Remaining two white balls are placed together in any one urn
  
  As seen in case 3, this gives another 10C6 × 4C4 × 10C1 possibilities
  
  Unlike what happened in case 1 and case 2, no overcounting occurs here
  
  Case 5 : Any of the 5 urns contain one white ball each. Rest of the 5 urns contain one black ball each. Remaining one white ball is placed in any of the 5 urns which already has a white ball. Remaining one black ball is placed in any of the 5 urns which already has a black ball.
  
  (Note that if we place the last white ball and/or the last black ball in urns where a different colour ball is already placed, it will result in overcounting of what we have already taken in case 1, case 2, case 3 and case 4)
  
  5 urns can be selected in 10C5 ways.
  Put one white ball in each of the selected 5 urns. 1 way of doing this
  Remaining 5 urns can be selected in 5C5 ways.
  Put one black ball in each of these 5 urns. 1 way of doing this
  Select any one urn which has a white ball in 5C1 ways
  place the remaining white ball in this urn. 1 way of doing this
  Select any one urn which has a black ball in 5C1 ways
  place the remaining black ball in this urn. 1 way of doing this
  
  Total number of ways = 10C5 × 5C5 × 5C1 × 5C1
  
  Therefore, required number of ways
  
  = (10C6×4C4×10C2)+(10C6×4C4×10C2)-(10C6×4C4×6C2)+(10C6×4C4×10C1)
  +(10C6×4C4×10C1)+(10C5×5C5×5C1×5C1)
  
  = (10C6×10C2)+(10C6×10C2)-(10C6×6C2)+(10C6×10C1)+(10C6×10C1)+(10C5×5C1×5C1)
  
  = 2(10C6×10C2)-(10C6×6C2)+2(10C6×10C1)+(10C5×5C1×5C1)
  
  = 2(210×45)-(210×15)+2(210×10)+(252×5×5)
  
  = 18900 - 3150 + 4200 + 6300
  = 26250
• 7 years agoHelpfull: Yes(0) No(0)
• 6 white and 6 black balls of same size are distributed among 10 urns so that there is atleast one ball in each urn. What are number of different distributions of balls?A) 25000. B)26250. C)28250. D)13125
  
  This question is useful (0) This question is not useful (0) Comment Answer
  permutations & combinations
  2 years ago, Rahul Rawat
  
  
  1 Answer
  Answer is 26250
  
  The following assumptions are taken for this answer
  a. All white balls are identical, all black balls are identical.
  b. All urns are distinct
  c. The order in which balls are placed inside the urn does not matter
  
  10 distinct urns are there, say 1,2,....10. Each urn must have at least one ball.
  
  Case 1 : Any of the 6 urns contain one white ball each. Rest of the 4 urns contain one black ball each. Remaining two black balls are distributed in any two urns.
  
  6 urns can be selected in 10C6 ways.
  Put one white ball in each of the selected 6 urns. Since the white balls are identical, this can be done only in 1 way
  4 urns can be selected from the remaining 4 urns in 4C4 ways .
  Put one black ball in each of the selected 4 urns. 1 way of doing this
  Select any two urns in 10C2 ways
  Place two black balls in these urns. 1 way of doing so
  
  Total number of ways = 10C6 × 4C4 × 10C2
  
  Case 2 : Any of the 6 urns contain one black ball each. Rest of the 4 urns contain one white ball each. Remaining two white balls are distributed in any two urns
  
  Similar to what we have seen in case 1, this can also be done in 10C6 × 4C4 × 10C2 ways
  
  But case 2 and case 1 together has resulted in overcounting of some of the possibilities . Suppose, in case 1, a white ball is placed in each of the 1-6 urns, a black ball is placed in each of the 7-10 urns and remaining two black ball is placed in 5 and 6, as given below
  
  1 2 3 4 5 6 7 8 9 10
  W W W W W W B B B B
  B B
  
  Assume, in case 2, a black ball is placed in each of the 5-10 urns, a white ball is placed in each of the 1-4 urns and remaining two white balls are placed in 5 and 6 urns,as given below
  
  1 2 3 4 5 6 7 8 9 10
  W W W W B B B B B B
  W W
  
  As we observe, such possibilities which are already counted in case1, have been counted in case 2 as well. This overcounting occurs in case 2, when the last two white balls are placed in urns which has a black ball. Count of such possibilities are 10C6 × 4C4 × 6C2. This need to be reduced from the total to balance the effect of overcounting.
  
  Case 3 : Any of the 6 urns contain one white ball each. Rest of the 4 urns contain one black ball each. Remaining two black balls are placed together in any one urn.
  
  6 urns can be selected in 10C6 ways.
  Put one white ball in each of the selected 6 urns. 1 way of doing this
  4 urns can be selected from the remaining 4 urns in 4C4 ways .
  Put one black ball in each of the selected 4 urns. 1 way of doing this
  Select any one urn in 10C1 ways
  Place two black balls in these urns. 1 way of doing so
  
  Total number of ways = 10C6 × 4C4 × 10C1
  
  Case 4 : Any of the 6 urns contain one black ball each. Rest of the 4 urns contain one white ball each. Remaining two white balls are placed together in any one urn
  
  As seen in case 3, this gives another 10C6 × 4C4 × 10C1 possibilities
  
  Unlike what happened in case 1 and case 2, no overcounting occurs here
  
  Case 5 : Any of the 5 urns contain one white ball each. Rest of the 5 urns contain one black ball each. Remaining one white ball is placed in any of the 5 urns which already has a white ball. Remaining one black ball is placed in any of the 5 urns which already has a black ball.
  
  (Note that if we place the last white ball and/or the last black ball in urns where a different colour ball is already placed, it will result in overcounting of what we have already taken in case 1, case 2, case 3 and case 4)
  
  5 urns can be selected in 10C5 ways.
  Put one white ball in each of the selected 5 urns. 1 way of doing this
  Remaining 5 urns can be selected in 5C5 ways.
  Put one black ball in each of these 5 urns. 1 way of doing this
  Select any one urn which has a white ball in 5C1 ways
  place the remaining white ball in this urn. 1 way of doing this
  Select any one urn which has a black ball in 5C1 ways
  place the remaining black ball in this urn. 1 way of doing this
  
  Total number of ways = 10C5 × 5C5 × 5C1 × 5C1
  
  Therefore, required number of ways
  
  = (10C6×4C4×10C2)+(10C6×4C4×10C2)-(10C6×4C4×6C2)+(10C6×4C4×10C1)
  +(10C6×4C4×10C1)+(10C5×5C5×5C1×5C1)
  
  = (10C6×10C2)+(10C6×10C2)-(10C6×6C2)+(10C6×10C1)+(10C6×10C1)+(10C5×5C1×5C1)
  
  = 2(10C6×10C2)-(10C6×6C2)+2(10C6×10C1)+(10C5×5C1×5C1)
  
  = 2(210×45)-(210×15)+2(210×10)+(252×5×5)
  
  = 18900 - 3150 + 4200 + 6300
  = 26250
• 7 years agoHelpfull: Yes(0) No(0)
• Answer is 26250
  
  The following assumptions are taken for this answer
  a. All white balls are identical, all black balls are identical.
  b. All urns are distinct
  c. The order in which balls are placed inside the urn does not matter
  
  10 distinct urns are there, say 1,2,....10. Each urn must have at least one ball.
  
  Case 1 : Any of the 6 urns contain one white ball each. Rest of the 4 urns contain one black ball each. Remaining two black balls are distributed in any two urns.
  
  6 urns can be selected in 10C6 ways.
  Put one white ball in each of the selected 6 urns. Since the white balls are identical, this can be done only in 1 way
  4 urns can be selected from the remaining 4 urns in 4C4 ways .
  Put one black ball in each of the selected 4 urns. 1 way of doing this
  Select any two urns in 10C2 ways
  Place two black balls in these urns. 1 way of doing so
  
  Total number of ways = 10C6 × 4C4 × 10C2
  
  Case 2 : Any of the 6 urns contain one black ball each. Rest of the 4 urns contain one white ball each. Remaining two white balls are distributed in any two urns
  
  Similar to what we have seen in case 1, this can also be done in 10C6 × 4C4 × 10C2 ways
  
  But case 2 and case 1 together has resulted in overcounting of some of the possibilities . Suppose, in case 1, a white ball is placed in each of the 1-6 urns, a black ball is placed in each of the 7-10 urns and remaining two black ball is placed in 5 and 6, as given below
  
  1 2 3 4 5 6 7 8 9 10
  W W W W W W B B B B
  B B
  
  Assume, in case 2, a black ball is placed in each of the 5-10 urns, a white ball is placed in each of the 1-4 urns and remaining two white balls are placed in 5 and 6 urns,as given below
  
  1 2 3 4 5 6 7 8 9 10
  W W W W B B B B B B
  W W
  
  As we observe, such possibilities which are already counted in case1, have been counted in case 2 as well. This overcounting occurs in case 2, when the last two white balls are placed in urns which has a black ball. Count of such possibilities are 10C6 × 4C4 × 6C2. This need to be reduced from the total to balance the effect of overcounting.
  
  Case 3 : Any of the 6 urns contain one white ball each. Rest of the 4 urns contain one black ball each. Remaining two black balls are placed together in any one urn.
  
  6 urns can be selected in 10C6 ways.
  Put one white ball in each of the selected 6 urns. 1 way of doing this
  4 urns can be selected from the remaining 4 urns in 4C4 ways .
  Put one black ball in each of the selected 4 urns. 1 way of doing this
  Select any one urn in 10C1 ways
  Place two black balls in these urns. 1 way of doing so
  
  Total number of ways = 10C6 × 4C4 × 10C1
  
  Case 4 : Any of the 6 urns contain one black ball each. Rest of the 4 urns contain one white ball each. Remaining two white balls are placed together in any one urn
  
  As seen in case 3, this gives another 10C6 × 4C4 × 10C1 possibilities
  
  Unlike what happened in case 1 and case 2, no overcounting occurs here
  
  Case 5 : Any of the 5 urns contain one white ball each. Rest of the 5 urns contain one black ball each. Remaining one white ball is placed in any of the 5 urns which already has a white ball. Remaining one black ball is placed in any of the 5 urns which already has a black ball.
  
  (Note that if we place the last white ball and/or the last black ball in urns where a different colour ball is already placed, it will result in overcounting of what we have already taken in case 1, case 2, case 3 and case 4)
  
  5 urns can be selected in 10C5 ways.
  Put one white ball in each of the selected 5 urns. 1 way of doing this
  Remaining 5 urns can be selected in 5C5 ways.
  Put one black ball in each of these 5 urns. 1 way of doing this
  Select any one urn which has a white ball in 5C1 ways
  place the remaining white ball in this urn. 1 way of doing this
  Select any one urn which has a black ball in 5C1 ways
  place the remaining black ball in this urn. 1 way of doing this
  
  Total number of ways = 10C5 × 5C5 × 5C1 × 5C1
  
  Therefore, required number of ways
  
  = (10C6×4C4×10C2)+(10C6×4C4×10C2)-(10C6×4C4×6C2)+(10C6×4C4×10C1)
  +(10C6×4C4×10C1)+(10C5×5C5×5C1×5C1)
  
  = (10C6×10C2)+(10C6×10C2)-(10C6×6C2)+(10C6×10C1)+(10C6×10C1)+(10C5×5C1×5C1)
  
  = 2(10C6×10C2)-(10C6×6C2)+2(10C6×10C1)+(10C5×5C1×5C1)
  
  = 2(210×45)-(210×15)+2(210×10)+(252×5×5)
  
  = 18900 - 3150 + 4200 + 6300
  = 26250
• 6 years agoHelpfull: Yes(0) No(1)