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Algebra
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5 Dec 2006 02:15:31 IST
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WE HAVE 30 BALLS AND WE HAVE TO DIVIDE THEM INTO THREE GROUPS.
THIS CAN BE DONE BY PLACING TWO MARKERS BETWEEN THEM.
. . . . .
No. of spaces = 29
Ways of placing two markers= 29C2
BUT SINCE THE BOXES ARE IDENTICAL SO TOTAL NO. OF CASES WII BE ONE-THIRD
= (29C2 )/3
THIS CAN BE DONE BY PLACING TWO MARKERS BETWEEN THEM.
. . . . .
No. of spaces = 29
Ways of placing two markers= 29C2
BUT SINCE THE BOXES ARE IDENTICAL SO TOTAL NO. OF CASES WII BE ONE-THIRD
= (29C2 )/3
6 Dec 2006 19:52:04 IST
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first off all supplying 3 balls to 3 boxes. Remaining are 27balls. Consider a box. It can collect 1,2,3,4,........27 balls with it. Hence 27 casses possible with a box. NO MATTER WHAT THE BOX IT IS since all boxes are identical. So the answere must be 27.















when balls as well as boxes are identical,the number of combinations and arrangements will be 1 each in 12 cases.
therefore, the required number of ways
=1x1x12=12 ways