Although your solution is correct.. the example that you gave is I think a typing mistake..sureshbala wrote:Once we fix the position of three digits in the correct position, the remaining three digits must be placed in the correct position and this can be done in only 2 ways.
For example, if we fix the digits 1,2 and 3 the arrangements possible are
1,2,3, 6.4,5
1,2,3, 6,5,4
i.e. once 1,2 and 3 are fixed 4 must be placed either in the 5th or the 6th position and for each of these two ways there will be only 1 way to fix the other 2 numbers.
Since we can fix the position of any 3 digits from 6 digits in 6C3 = 20 ways, total number of arrangements = 20 x 2 =40
Yup...that's a typo...thanks for the correctiondumb.doofus wrote:Although your solution is correct.. the example that you gave is I think a typing mistake..sureshbala wrote:Once we fix the position of three digits in the correct position, the remaining three digits must be placed in the correct position and this can be done in only 2 ways.
For example, if we fix the digits 1,2 and 3 the arrangements possible are
1,2,3, 6.4,5
1,2,3, 6,5,4
i.e. once 1,2 and 3 are fixed 4 must be placed either in the 5th or the 6th position and for each of these two ways there will be only 1 way to fix the other 2 numbers.
Since we can fix the position of any 3 digits from 6 digits in 6C3 = 20 ways, total number of arrangements = 20 x 2 =40
1) 1,2,3,6,4,5
2) 1,2,3,5,6,4 (This one your typed incorrectly)
