find the total number of combinations of 5 letters a,b,a,b,b taking some or all at a time?

• total no of ways of taking some or all out of p+q things such that p are of one type and q are of another type is (p+1)(q+1)-1
  so,ans is a comes 2 times i.e. p=2
  b comes 3 times i.e q=3
  ans =(2+1)(3+1)-1
  =11
• 11 years agoHelpfull: Yes(54) No(5)
• first 1 digit no. can be a or b(2 ways 2 select), for 2 digit no.fallowing comb.
  r possible..aa,ab,ba,bb(4ways)
  for 3 digit no's lets 3b,2b nd 1a,1b nd 2a.
  so total ways 1+3+3=7ways.
  for 4 digit no's 3b nd 1a,2b nd 2a,
  so total ways 4+6=10ways(coz using 3b nd 1a..4!/3!=4 nd 4!/2!*2!=6).
  for 5 digit no.s 3b nd 2a.
  so total ways=5!/3!*2!=10
  so total combination of no.s(1digit+2digit+3digit+4digit+5digit)..
  2+4+7+10+10=33
  so ans will be 33.
• 11 years agoHelpfull: Yes(20) No(21)
• 5!/(2!*3!)=10.
  so! 10 is d ri8 ans.
• 11 years agoHelpfull: Yes(8) No(5)
• 1 letter can be chosen in 2 ways. a or b
  2 letters can be chosen in 3 way. aa, ab, bb
  3 letters can be chosen in 3 ways. bbb, aab, bba
  4 letters can be chosen in 2 ways. aabb, bbba
  5 letters can be chosen in 1 way.
  So total ways are 11
  
• 9 years agoHelpfull: Yes(6) No(1)
• 5!/(2!*3!)=20
• 11 years agoHelpfull: Yes(5) No(15)
• if we consider a , b as a factors then the question reflects as factors of a given number
  so we get a^2*b^3
  no of factors we can get from formula (2+1)*(3+1)=12
  in factors 1 is also a factor .. here there is no such case (i.e not selecting any)
  so 12-1 = 11 will be the answer
  
• 11 years agoHelpfull: Yes(4) No(1)
• The idea is to find not the combinations but the freq of letters.
  ababb
  2a's 3b's 1 letter stuffs a,b=2,2 letter stuffs aa,bb,ab=3,3letter stuffs aab,bbb,abb=3,4 letter stuffs=abbb,aabb=2,5 letter stuffs=1 adding these up we get 2+3+3+2+1= 11ways
• 9 years agoHelpfull: Yes(2) No(0)
• 11ways is correct answer
• 9 years agoHelpfull: Yes(2) No(0)
• 1 letter can be chosen in 2 ways. a or b
  2 letters can be chosen in 3 way. aa, ab, bb
  3 letters can be chosen in 3 ways. bbb, aab, bba
  4 letters can be chosen in 2 ways. aabb, bbba
  5 letters can be chosen in 1 way.
  So total ways are 11
• 8 years agoHelpfull: Yes(2) No(0)
• (5c1+5c2+5c3+5c4+5c5)(3!*2!)=31
• 11 years agoHelpfull: Yes(1) No(8)
• correct ans given is 11..but dnt know how to solve
• 11 years agoHelpfull: Yes(1) No(10)
• Ans is 11
  The best and detailed answer was given here..
  http://linkshrink.net/7dq5SJ
• 9 years agoHelpfull: Yes(1) No(7)
• I think exact ans is 15.
  5p1+5p5/2!*3!
  so by solving it ans is 15.
• 11 years agoHelpfull: Yes(0) No(7)
• i think it should be 10
  
• 8 years agoHelpfull: Yes(0) No(0)
• 1 letter can be chosen in 2 ways. a or b
  2 letters can be chosen in 3 way. aa, ab, bb
  3 letters can be chosen in 3 ways. bbb, aab, bba
  4 letters can be chosen in 2 ways. aabb, bbba
  5 letters can be chosen in 1 way.
  So total ways are 11
• 8 years agoHelpfull: Yes(0) No(1)