{"id":6769,"date":"2021-10-21T14:13:49","date_gmt":"2021-10-21T08:43:49","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6769"},"modified":"2021-10-25T01:33:30","modified_gmt":"2021-10-24T20:03:30","slug":"how-many-numbers-can-be-formed-with-the-digits-1-2-3-4-3-2-1-so-that-the-odd-digits-always-occupy-the-odd-places","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/how-many-numbers-can-be-formed-with-the-digits-1-2-3-4-3-2-1-so-that-the-odd-digits-always-occupy-the-odd-places\/","title":{"rendered":"How many numbers can be formed with the digits 1, 2, 3, 4, 3, 2, 1 so that the odd digits always occupy the odd places?"},"content":{"rendered":"
There are 4 odd digits (1, 1, 3, 3) and 4 odd places(first, third, fifth and seventh).<\/p>\n
At these places the odd digits can be arranged in \\(4!\\over 2!2!\\) = 6<\/p>\n
Then at the remaining 3 places, the remaining three digits(2, 2, 4) can be arranged in \\(3!\\over 2!\\) = 3 ways<\/p>\n
Therefore,\u00a0 The required number of numbers = 6 \\(\\times\\) 3 = 18<\/p>\n
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