Permutations and Combinations

Given, word Probability
Number of letter in probability
P,R,O,A,L,T,Y,BB,II
(i) 4 letter are selected alone different
i.e. ‌9C4=‌

9!

4!5!

=126
(ii) Two letter or alike other are different
2C1×8C2=‌

2×8!

2!6!

=56
(iii) Two letter are alike
2C2=1
Total number of sample space
=126+56+1=183
Favorable case in which at least one letter is repeated
‌ i.e., ‌‌‌56+1=57
∴‌‌ Reqmend probability =‌

57

183

=‌

19

61