there are 5 dogs and three cats. what is the probability that there is one cat at both the ends?

• (3C2*2!*6!)/8!=3/28
• 9 years agoHelpfull: Yes(24) No(1)
• Let Dog= D and Cat= C
  then the arrangement will be like CD_D_D_D_DC if their is Cat at both ends. Remaining 1 Cat cat be fit any of this gap. so 4C1 = 4
  now 5dogs and 3cats can be arranged 6C3= 20
  the probability P(E)=4/20 = 1/5
• 9 years agoHelpfull: Yes(5) No(7)
• @vikash kumar explain in detail
• 9 years agoHelpfull: Yes(1) No(1)
• suppose we have 8 places to fill and at 1st position and last position we have select 2 cats out of 3..
  now at first place we will be selecting 1 cat out of 3=3C1 ..so ultimately at last we have to choose 1 cat out of 2 cats=2C1...now we are left with 5 dogs and 1 cat..possible way of arranging all 6 =6!..between two cars at extremeties...
  total arrangements=8!..
  so answer is 3C1*6!*2C1/8!=3/28
• 5 years agoHelpfull: Yes(1) No(0)