In a family 7 children dont eat spinach, 6 dont eat carrot, 5 dont eat beans, 4 dont eat spinach & carrots, 3 dont eat carrot & beans, 2 dont eat beans & spinach. One doesnt eat all 3. Find the no. of children.

• since,n(aUbUc)=n(a)+n(b)+n(c)-n(a intersection b)-n(a intersection c)-n(b intersection c)+n(a intersection b intersection c)
  therfore,n(aubuc)=7+6+5-4-3-2+1=10
• 13 years agoHelpfull: Yes(12) No(0)
• do not spinach only=2
  do not carrot only=0
  do not beans only=1
  do not spinach & carrots only=3
  do not beans & carrots only=2
  do not spinach & beans only=1
  do not all three=1
  total 10 children(2+0+1+3+2+1+1)
  
• 13 years agoHelpfull: Yes(4) No(2)
• 10 children
• 14 years agoHelpfull: Yes(1) No(1)
• S: Children who don't eat spinach
  C: " " " " carrot
  B: " " " " beans
  S=7, C=6, B=5;
  S^C=4 ; C^B= 3; B^S=2; S^C^B=1; (where '^' implies intersection & U - union)
  SUBUC= S + C + B - (S^C) -(S^B) -(B^C) + (S^B^C)
  SUBUC= 7 + 6 + 5 - 4 - 3 - 2 + 1
  SUBUC= 10
• 13 years agoHelpfull: Yes(1) No(0)