Heya ❤️❤️❤️❤️❤️
HELP PLEASE ❤️❤️
✒️ In a village,there are 87 families of which 52 have atmost 2 children. In a rural development programme, 20 families are to be helped choosen for assistance, of which atleast 18 families must have 2 children. In how many ways can the choice be made ?❓❓❔
NO SPAM ❌❌
NO COPY ❌❎
A QUALITY ANSWER REQUIRED ✔️☺️☺️☺️☺️☺️
Answers
Answered by
9
HERE'S YOUR ANSWER
⬇️⬇️⬇️⬇️⬇️⬇️⬇️⬇️
Given ,
There are total 87 families in a village.
Among them , 52 have atmost 2 Children.
✒️Now , 87-52 = 35 don't have atmost 2 children.
20 must be choosen of which 18 must have 2 children
So , • 52^C18× 35^C2
=( 52!/18!×34!)(35!/2!×33!)
✒️Again , 19 out of 52 and 1 out of 35
So , • 52C19× 35C1
= (52!/19!×33!)(35!/1!×34!)
✒️ And 20 families must be choosen from 52 so ,
• 52C20
= 52!/20!×32!
Therefore total combination
= (52!/18!×34)(35!/2!×33!)
+(52!/19!×33!)(35!/1!×34!)+(52!/20!×32!)
HOPE WILL HELP ✔️✔️✔️✔️
⬇️⬇️⬇️⬇️⬇️⬇️⬇️⬇️
Given ,
There are total 87 families in a village.
Among them , 52 have atmost 2 Children.
✒️Now , 87-52 = 35 don't have atmost 2 children.
20 must be choosen of which 18 must have 2 children
So , • 52^C18× 35^C2
=( 52!/18!×34!)(35!/2!×33!)
✒️Again , 19 out of 52 and 1 out of 35
So , • 52C19× 35C1
= (52!/19!×33!)(35!/1!×34!)
✒️ And 20 families must be choosen from 52 so ,
• 52C20
= 52!/20!×32!
Therefore total combination
= (52!/18!×34)(35!/2!×33!)
+(52!/19!×33!)(35!/1!×34!)+(52!/20!×32!)
HOPE WILL HELP ✔️✔️✔️✔️
Answered by
7
Hey Dear !
=) Please , Refer the attachment .
Hope , It Helps .
Attachments:
Anonymous:
thanx
bro
welcm bro :-)
Similar questions