How many 4 letter words can be formed from abcd with at least 2 letters repeated?
Answers
Answer:
Letters:- C-1 E-3 F-2 I-2 N-1 T-1 V-1
ABCD type words : 7C4×4!
ABCC type words : 3C1×6C2×4!2!
ABBB type words : 1×6C1×4!3!
AABB type words : 3C2×4!2!×2!
Sum them over to get the total!
NOTE: AABB (for example) means all words (including permutations) with 2 letters of one kind and two letters of another type.
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answered
Dec 3 '16 at 12:42
There are 11 letters in 'INEFFECTIVE'.
CASE (1): There is only set of three same letters of the letter E. So 3 same letters can be selected in one way. Out of the six remaining distinct letters we can select one in (61)=6 ways. Hence there are 6 groups each of which contains the three same letters. These four letters can be arranged in (43)=4 ways. Hence the total number of words with 3 same and one different letters is 6×4=24 ways.
CASE (2): There are 3 sets of two same letters E,F,I. Out of these 3 sets, two can be selected in (32)=3 ways. So there are 3 groups each of which contains 4 letters two of which are same of one type and the other two of a different same type. Now 4 letters in each group can be arranged in (42)=6 ways. Hence the total number of words with two letters of one kind and the other two of the different kind is 3×6=18 times.
CASE (3): Out of 3 sets of 2 same letters one set can be chosen in (31)=3 ways. Now from the remaining 6 letters, 2 distinct letters can be chosen in (62)=15 ways. So there are 3×15=45 groups of 4 letters each. Now letters of each group can be arranged in 4!2!=12 ways. Hence, the total number of ways consisting of two same letters and 2 different are 45×12=540 ways.
CASE (4): There are (74) groups of words with 4 different letters each. In each group each letter can be arranged in 4!=24 ways. So the total number of 4 letter words in which all letters are different is (74)×24=840 ways.
There are thus a total of 24+18+540+840=1422
Step-by-step explanation:
For atleast 2 letters to be repeated we should have 2 pair each
2 alike
4c1×6c2×12
2 alike 2 alike
4c1×3c1×6
For 3 alike we should have 3 pair each
3 alike
4c1×9c1×4
For 4 alike we should have 4 pair each
4c1
therefore ans= 940
here you should give equality so i took equal pair when reptetions