Question
Out of 10 men, 3 are wearing red caps, 4 are wearing green caps, and the remaining men are wearing orange caps. They are to be seated in a linear row. If the number of ways they can be arranged such that no more than three men wearing green caps sit next to each other is 'q', find the sum of the squares of each digit of 'q'.
Solution
ATQ,
Total number of ways in which all of them can seat = 10! = 36,28,800 Number of ways in which all 4 men wearing green cap sit next to each other:
= 7! X 4! = 1,20,960
'q' = 36,28,800 - 1,20,960 = 35,07,840
Required sum = 3² + 5² + 0² + 7² + 8² + 4² + 0²
= 9 + 25 + 0 + 49 + 64 + 16 + 0 = 163
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