Badiuddin askIITians.ismu Expert
Last Activity: 14 Years ago
Dear manisha
here letters are 4 i , 2 n , f , t , e, s , m, a, l
case 1: alll 5 are distinct
number if words = 9C5 5!
case 2: 2 same and 3 distinct
number of words =2C1 * 8C3 * 5!/2!
case 3 : 2 alike of one kind + 2 laike of one kind + 1 distinct
number of words = 2C2 * 7C1 *5!/2! 2!
case 4: 3 alike of one kind + 2 alike of other kind
number of words = 1 *1 *5!/3!2!
case 5 : 3 alike of one kind and 2 distict
number of words = 1 * 8C2 * 5!/3!
so add all above u will get the desired result
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