A four-digit number is formed by using only the digits 1, 2 and 3 such that both 2 and 3 appear at least once. The number of all such four-digit numbers is
Explanation:
Case 1: Number contains 2, 3, 1, 1 Number of such numbers = 4!/2! = 12
Case 2: Number contains 2, 3, 3, 1 Number of such numbers = 4!/2! = 12
Case 3: Number contains 2, 2, 3, 1 Number of such numbers = 4!/2! = 12
Case 4: Number contains 2, 2, 3, 3 Number of such numbers = 4!/(2!×2!) = 6
Case 5: Number contains 2, 2, 2, 3 Number of such numbers = 4!/3! = 4
Case 6: Number contains 2, 3, 3, 3 Number of such numbers = 4!/3! = 4
∴Total such numbers = 12 + 12 + 12 + 6 + 4 + 4 = 50.
Hence, 50.
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