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There are 8 digits 1, 2, 0, 2, 4, 1, 2, 4 in which 1 occurs 2 times, 2 occurs 3 times and 4 occurs 2 times ∴ Number of 8 digit numbers = 8!/(2! ×3! ×2!) = 1680 But out of these 1680 numbers there are some numbers which begin with ‘0’ and they are not 8 digit numbers The numbers of such numbers beginning with ‘0’ = 7!/(2! ×3! ×2!) = 210 Hence the required number of 8 digit numbers = 1680 – 210 = 1470
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