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![[Post New]](/templates/default/images/icon_minipost_new.gif) 22 Mar 2008 10:35:42 IST
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PROPORTION
P - 2
R - 2
O - 3
T - 1
I - 1
N - 1
So 4 lettered word
cases :
4 different , 3 alike 1 diff , 2 alike 2 diff , 2 alike 2 alike
case 1
4 diff : 6 P 4 = 360
case 2
3 alike 1 diff
3 O , 2 Ps , 2 R s so it can be arranged in 4! / 3! = 4 ways remaining 1 letter can be filled in 5 ways
So number of ways = 5*4 = 20
case 3
2 alike 2 diff
repeating letters are P,R,O
so they can be arranged in 3 ways .
Hence total arrangement = 3 *5 C2*4!/2! = 360
case 4
2 alike 2 alike No of ways = 3 C2 * 4!/(2!*2!) = 18
So total = 758
Koni sir  please provide the multinomial theorem solution
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this reply: 15 points
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