Question

how many 4 letter words can be formed by using the letters of the word ineffective

Answer 1

11C4=11!/4!7!=(11*10*9*8)/(4*3*2*1)=330

hope this helps:p

Answer 2

Hi friend,

There are 11 letters in the word INEFFECTIVE i.e E E E, F F, I I, C, T, N, V=6c1

(i) There is only one set of three same letters i.e E E E.These 4 letter can be arranged in ways=4!/3!1!

Hence the total number of words consisting of 3 same and one different letters = 6c1×4!
6×4=24.



(ii) There are three sets of two some letters i.e E E, F F, I I . Out of these three sets two can be selected in ways.Now, 4 letters in each group can be arranged in ways.=3c2

Hence the total number of words consisting two same letters of one kind and 2 same letters of other kind =3c2×4!/2!2!
=3×6=18

(iii) Out of 3 sets of two same letters one set can be chosen in ways.Now, from the remaining 6 distinct letters, 2 letters can be chosen in ways. So there are group of 4 letters each

=3c1×6c2

Now, letters of each group can be arranged among themselves in ways. = 4!/2!

Hence, the total number of words consisting two same letters and 2 different = (3c1×6c2)×4!/2!
=3×15×6=270

(iv) There are 7 distinct letters E, F.So the total number of 4 letters words in which all letters are different = 7c4×4!
=840.....

Hope this helps you...
Please mark it as brainliest answer...☺☺☺