Answer:

There are 5 distinct LETTERS (B,A,R,C & K) in the word BARRACK with A &R being REPEATED twice. To form 4-lettered words, there will be 4 situations viz. (i) all 4 different (ii) 2 are different and other 2 are A's (iii) 2 are different and other 2 are R's (iv) 2 are A's and other 2 are R's.

Case (i): The first letter out of 5 can be chosen in 5 ways. The second letter can be chosen out of REMAINING 4 in 4 ways. Again, the third letter can be chosen out of remaining 3 in 3 ways and lastly the FOURTH letter can be chosen out of remaining 2 in 2 ways. Hence, total arrangements of 4 lettered words with no repetition is 5*4*3*2 = 120.

Case (ii): Here, 2 letters are different and other 2 letters are both A . But, 2 letters can be chosen out of 4 ( B,R,C & K) in C(4,2)=6 ways. Again, this 4-lettered word with 2 A 's can be arranged in 4!/2!=12 ways. So, total arrangement is 6*12 =72.

Case (iii) : Similarly, when 2 letters are different and 2 letters are both R , the total arrangement is also 72.

Case (iv) : Here, 2 letters are both R and remaining 2 letters are both A and hence, the total arrangement is 4!/(2!∗2!)=6.

So, the total number of 4 lettered words is 120+72+72+6 = 270