document.write( "Question 1184895: If a permutation is chosen random from the letters “aaabbbccc” what is the probability that it begins with at least 2 a’s \n" ); document.write( "

Algebra.Com's Answer #815561 by greenestamps(13200) $\"\"$ $\"About$

You can put this solution on YOUR website!

\n" ); document.write( "(1) Total number of permutations:

\n" ); document.write( " $\"%289%21%29%2F%28%283%21%29%283%21%29%283%21%29%29=1680\"$

\n" ); document.write( "(2) Number of permutations starting with all three a's:

\n" ); document.write( " $\"%286%21%29%2F%28%283%21%29%283%21%29%29+=+720%2F36=20\"$

\n" ); document.write( "(3) Number of permutations starting with exactly two a's:

\n" ); document.write( " $\"%287%21%29%2F%281%21%29%283%21%29%283%21%29+=+5040%2F36+=+140\"$

\n" ); document.write( " $\"P=%2820%2B140%29%2F1680=2%2F21\"$

\n" ); document.write( "ANSWER: 2/21

\n" ); document.write( "----------------------------------------------------------------

\n" ); document.write( "Tutor @ikleyn is right -- I double counted some of the permutations.

\n" ); document.write( "The 20 permutations in (2) above are counted again in (3); the number of permutations starting with at least 2 a's is 140, as shown in (3).

\n" ); document.write( "Then the probability of a permutation starting with at least 2 a's is 140/1680 = 1/12, which agrees with her answer.

\n" ); document.write( " \n" ); document.write( "

\n" );