Combination

Answer is none of these Consider Examination It has letters E, X, A, M, I, N, T, O where A, I, N are repeated twice. So, four letter word may consist of (i) 2 alike letters of one kind and 2 alike of letter of the second kind. (ii) All different (iii) 2 alike and 2 distinct letter So in (i) There are three sets of alike letters, namely AA, NN, II. Out of these three sets, two can be selected in 3C2 ways. So, there are 3C2 groups, each containing 4 letters out of which two are alike letters of one kind and two are like letters of the second kind. Now, 4 letters in each group can be arranged in 4!2!×2! Therefore, the total number of four letters words are 3C2×4!2!×2!=3×6=18 Since, The total number of letter =11 Similar way, (ii) All different i.e. 8C4×4!=8!4! As selecting 4 out of 8 different letters is 8C4 Arrangement of these 4 are done by 4! Thus, 8!4!=8×7×6×5×4!4!=56×30=1680 (iii) 2 alike and 2 distinct letter: 3C1×7C2=3×7×6×5!5!×2=3×7×62=378 [ Since, nCr=nCn−r] ∴ Total number of ways in which 4 letter words are formed = 1680 +378 +18 +378=2454 ways. Consider PROPORTION We have P & R 2 times, O 3 times and other letters T, I, and N once 1. Words with four distinct letters. We have 6 letters all total (I,N,P,R,O and T) so we can arrange this letter in 6 C4 ×4!=360 ways 2. Words with exactly a letter repeated twice. We have P, R and O repeating itself. Now one of this three letter can be choose in 3 C1 =3 ways The other two distinct letters can be selected in 5 C2 =10 ways. Now each combination can be arranged in 2!4! =12 ways. So total number of such words 3×10×12= 360 ​ 3. Words with exactly two distinct letters repeated twice Two letters out of the three repeating letters P, R and O can be selected in 3 C2 =3 ways. Now each combination can be arranged in 2!×2!4! =6 So, total number of such words =3×6= 18 ​ 4. Words with exactly a letter repeated thrice We have one portion for this as our main letter that is O. Now we have to select 1 letter out of the 5 remaining options so number of ways to select this 5 C1 =5 ways Now each combination can be arranged in 3!4! =4 So, total number of such words =5×4= 20 ​So, all possible number of arrangements =360+360+18+20=758 ways

Sudha Reddy

CA Final

★ 20K+

09-Aug-23 10:28