Answer on Question#38642 – Math - Other
We have L={w∣w has equal number of a’s, b’s, c’s or equal number a’s, b’s and d’s}L = \{w \mid w \text{ has equal number of a's, b's, c's or equal number a's, b's and d's}\}L={w∣w has equal number of a’s, b’s, c’s or equal number a’s, b’s and d’s}.
We know that context-free languages are not closed under intersection. So A is not context-free.
The correct answer is C.
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