Question

3. (20) In how many ways can the word KUTKUT be arranged so that no two identical letters are next to each other?

Answer 1

= 5! Now, KUTKUT is a 6 letter world with qk's, a T's and 2U's. KIT, U are So, Total no. of ways. repeating two 2!2!2! times so we divide Lby 21 three times ] No. of ways in which 2 K's are together we write ak's are together and the 2121 remaining words U,T,UT (4 words) Mly no. of ways in which 2 T's are Total words = 5 together = 51 as we consisder 2K to be a single word & diride by 21 twice Illy no. of ways in which 2 U's are as a T's & 2 U's are together = 51 - repeating ] 221 no. of ways in which ak's and I's are together (KKTT uu)! I we consider [KR) as one letter TT as me letter and u, vare 2 letters so total 4 letters as us are repeating twotime So, no. of ways = 4 So divide bu by 24 ] 21

noi of ways noi of ways when ak's 20's are together and TIT are and ther 2 world [KK QU TT So, no. of ways = 4! lly no. of ways au's and 2 T's are together and Kg Kare two words so, no. of ways = 2 Now, no, of ways when ak's , 2T's and au's are to together [KG ITT NO , so they are 3 words, So, no. of ways = 31 Now, let A= when 2k's are together B = when at's are together c = when zu's are together Now, n(AUBUC) = n(A) + n(B) +0(c) - n(ANB) - n(BOC) - n(ANC) + n(ANBOS) no. of ways in which atleast one of = 5 = +51 +51 - 41 -41 -41 +31 21 21 21 212 21 49 al al al = 3.51 -3.41 +31 21 21 Kuort together. So, Total no. of ways in which no identical letters are together = Total no. of ways - no. of ways in which ne (kuorT) is together

= 6! 2121 21 – 3,5! 21 21 + 3.41 21 – 3) 6x5X 4x3x2 2x2x2 – 3xSxyx3x2 + 3x4x342-6. xxx So o = 96 - 96 + 36-6 = 30 no. of w no. of ways in which no two identical are together is 30. letters