A) 700
B) 750
C) 758
D) 800
Correct Answer: C
Solution :
We have got \[2{{P}^{s}},\ 2{{R}^{s}},\ 3{{O}^{s}},\ 1I,\ 1T,\ 1N\ i.e.\]6 types of letters. We have to form words of 4 letters. We consider four cases (i) All 4 different: Selection \[^{6}{{C}_{4}}=15\] Arrangement \[=15\ .\ 4\ !\ =15\times 25=360\] (ii) Two different and two alike : \[{{P}^{s}},\ {{R}^{s}}\] and \[{{O}^{s}}\] in \[^{3}{{C}_{1}}=3\] ways. Having chosen one pair we have to choose 2 different letters out of the remaining 5 different letters in \[^{5}{{C}_{2}}=10\] ways. Hence the number of selections is \[10\times 3=30\]. Each of the above 30 selections has 4 letters out of which 2 are alike and they can be arranged in \[\frac{4\ !}{2\ !}=12\] ways. Hence number of arrangements is \[12\times 30=360\]. (iii) 2 like of one kind and 2 of other : Out of these sets of three like letters we can choose 2 sets in \[{{10}^{5}}-252=99748\] ways. Each such selection will consist of 4 letters out of which 2 are alike of one kind, 2 of the other. They can be arranged in \[\frac{4\ !}{2\ !\ 2\ !}=6\] ways. Hence the number of arrangements is \[3\times 6=18\]. (iv) 3 alike and 1 different : There is only one set consisting of 3 like letters and it can be chosen in 1 way. The remaining one letter can be chosen out of the remaining 5 types of letters in 5 ways. Hence the number of selection \[=5\times 1\]. Each consists of 4 letters out of which 3 are alike and each of them can be arranged in \[\frac{4\ !}{3\ !}=4\] ways. Hence the number of arrangements is \[5\times 4=20\]. From (i), (ii), (iii) and (iv), we get Number of selections \[=15+30+3+5=53\] Number of arrangements \[=360+360+18+20=758\].
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