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Permutation and Combination problem A 4-character Personal Identification Number (PIN) can be formed by 26 alphabets from A to Z and 10 digits from 0 to 9. How many different PINs can be formed if the Pin must contain at least one alphabet and no character is allowed to repeat itself? A. 58695 B. 58905 C. 1408680 D. 1413720 Please help and explain! thanks!
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total 36 characters. if no restriction, the permutation = 36P4 Now because alphabets must be here. so we see how many permutations are without any alphabets = 10P4 So answer = 36P4 - 10P4 = 1408680
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