The letters of the word calcutta and america are arranged in all possible ways. The ratio of the number of these arrangements is
Answers
Calcutta has 8 letters, but there are 2 "C"s, 2"A"s and 2"T"s.
If there were no duplicated letters, then there would be 8! ie 8x7x6x5x4x3x2x1 ways to arrange the letters.
The first could be any of 8
The second could be any of 7 (the first is already in place)
The third could be any of 6 (the 1st and 2nd are already in place)
and so on = 8x7x6x5x4x3x2x1 =40320 ways
However we could swop each position where one "C" is for the other "C" and vice versa. Therefore the number of combinations is double what it should be in respect of having two "C"s.
Similarly we need to divide by 2 for the "A"s and another divide by 2 for the 2 "T"s.
Therefore Calcutta has 8! / 2*2*2 = 8x7x6x5x4x3x2x1 / 8 = 7!
There are 5040 combinations for the letters of Calcutta
In a similar way America has 7 letters and 2 are "A"s.
Therefore the number of combinations for America = 7! / 2 = 2520 combinations
Comparing the two: Calcutta has 7! combinations and America has 7!/2 combinations
The ratio is therefore 5040 : 2520 or put another way 7! : 7!/2
Multiply both sides by 2 and divide both sides by 7!
Calcutta : America = 2 : 1