how many anagrams can you make from MATHEMATICS?
Answers
Answered by
2
i though 8!/2! anagrams
dreamsaalolika:
could you show the steps too
Answered by
2
this question's answer is 8!/2!
THEORY:
Permutations of nn things of which P1P1 are alike of one kind, P2P2 are alike of second kind, P3P3 are alike of third kind ... PrPr are alike of rthrth kind such that: P1+P2+P3+..+Pr=nP1+P2+P3+..+Pr=n is:
n!P1!∗P2!∗P3!∗...∗Pr!n!P1!∗P2!∗P3!∗...∗Pr!.
For example number of permutation of the letters of the word "gmatclub" is 8! as there are 8 DISTINCT letters in this word.
Number of permutation of the letters of the word "google" is 6!2!2!6!2!2!, as there are 6 letters out of which "g" and "o" are represented twice.
Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be 9!4!3!2!
THEORY:
Permutations of nn things of which P1P1 are alike of one kind, P2P2 are alike of second kind, P3P3 are alike of third kind ... PrPr are alike of rthrth kind such that: P1+P2+P3+..+Pr=nP1+P2+P3+..+Pr=n is:
n!P1!∗P2!∗P3!∗...∗Pr!n!P1!∗P2!∗P3!∗...∗Pr!.
For example number of permutation of the letters of the word "gmatclub" is 8! as there are 8 DISTINCT letters in this word.
Number of permutation of the letters of the word "google" is 6!2!2!6!2!2!, as there are 6 letters out of which "g" and "o" are represented twice.
Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be 9!4!3!2!
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