In how many ways can(2n + 1) identical balls be placed in 3 distinct boxes so that any two boxes together will contain more balls than the third box.
Text Solution
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Sol. x 1 + x 2 + x 3 = 2n + 1
Total number of ways of places = 2n + 3 C 2
Total number of ways to placed the balls so that first box have more balls than other two = n + 2 C 2
(first place (n + 1) balls in first box and then divide n balls in 3 boxes)
Hence total number of ways
2n + 3 C 2 – 3 n+ 2 C 2 = 
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