Question
In a game of twist, there are 10 indistinguishable coins which need to be arranged in the boxes below such that no row is empty and each box contains no more than one coin. In how ways can the coins be arranged in the following layout
Answer: Option A
:
A
Total number of possible selection of boxes for the 10 coins = 12C10=66
Cases in which rows will be empty = 2, when row 2, 3 and 4 and completely filled and when row 1,2, and 3 are completely filled. In all other cases, atleast one coin will be in each row.
Answer = 66 - 2 = 64
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:
A
Total number of possible selection of boxes for the 10 coins = 12C10=66
Cases in which rows will be empty = 2, when row 2, 3 and 4 and completely filled and when row 1,2, and 3 are completely filled. In all other cases, atleast one coin will be in each row.
Answer = 66 - 2 = 64
Was this answer helpful ?
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