Six different pencils and three bags are taken. Each pencil is to be put in one of these bags. No bag should remain empty and all bags should not have same number of pencils. In how many ways can this be done?
Solution
Correct Answer: Option C
Each pencil can be put in one of three bags.
⇒ Number of ways in which this can be done = 3 × 3 × 3 × 3 × 3 × 3 = 729
Out of these, there can be some cases when one or more bags are empty, and when all bags have same number of pencils.
Case i) one bag is empty
Empty bag can be selected in 3c1 ways i.e. 3 ways
Now, each pencil can be put in one of two ways.
Number of ways in which this can be done = 3 × (2 × 2 × 2 × 2 × 2 × 2) = 192
Case ii) two bags are empty
All pencils will be in one of three bags. This can be done in only 3 ways.
Case iii) three bags are empty. This case is not possible
Case iv) all bags have same number of pencils, i.e., 2 pencils.
⇒ We have to select two pencils for each bag.
Number of ways in which this can be done = 6c2 × 4c2 × 2c2 = 15 × 6 × 1 = 90
∴ Total number of ways in which this can be done = 729 – 192 – 3 – 90 = 444
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