Anonymous asked in Science & MathematicsMathematics · 1 month ago

In how many ways can you put 3 balls, colored red,white and blue in 20 empty boxes if the red and white are to be placed together.?

2 Answers

  • 1 month ago
    Favorite Answer

    It's just picking two boxes out of 20, one blue and another for both red and white.

    20 * 20 = 400, if all 3 balls in the same box is allowed, else

    20 * 19 = 380, if not.

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  • 1 month ago

    I'd say 648.

    I suppose that each box can hold only one ball and that empty boxes are in line, not in a circle or some other shape :)

    Number the boxes from left to right 1, 2, 3, ... 20.

    Place red ball in box 1 and white ball in box 2. Observe them as a linked pair where red in always on the left and white is on the right (remember this). There remains 18 empty boxes (3, 4, ..., 20). Blue ball can be placed in those boxes in 18 ways.

    Now move red and white together one place to the right, so that red is in 2 and white is in 3. Now again blue can be placed in 18 ways (1, 4, 5, ..., 20).

    Repeat this 18 times until red is in box 19 and white is in 20.

    So we had 18x18=  324 combinations.

    Remember that in a pair red ball was always left and white was right?

    Now return to the beginning and switch red and white so that white ball is in box 1 and red is in 2. Repeat the procedure to get 324 new combinations.

    324+324= 648.

    You can also write this as 2 x 18 x 18 = 648

    • husoski
      Lv 7
      1 month agoReport

      I read the problem differently, with "together" meaning "in the same box". Your analysis is right on the money for one ball per box.  Thumbs way up!

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