The number of ways in which six different prizes can be distributed among three children each receiving at least one prize is-

270

No worries! We‘ve got your back. Try BYJU‘S free classes today!

540

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1080

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2160

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $540$
Prizes can be distributed number of $1, 2, 3$ or $2, 2, 2,$ or $1, 4, 1$ to $3$ children $.$
Total no $.$ of ways $= \frac{6!}{3! 2!} \times 3! + \frac{6!}{2! 2! 2! 3!} \times 3! + \frac{6!}{4! 2!} \times 31$
$= 360 + 90 + 90$
$= 540$
Hence $,$ option $(B)$ is correct $.$

Suggest Corrections

Similar questions

The number of ways in which n different prizes can be distributed among m(< n) persons if each is entitled to receive at most n 1 prizes, is

Q. The number of ways in which

5

different prizes can be given among

11

candidates, each candidate receiving any number of prizes are

The number of ways in which n different prizes can be distributed among m (<n) persons if each is entitled to receive at most n - 1 prizes, is_______.

Q. The numbers of ways in which

n

prizes can be distributed among

n

students when each student is eligible to get any number of prizes is:

Q. Number of ways in which 4 prizes can be distributed among 5 student if no student gets more than one prize is

Join BYJU'S Learning Program

The number of ways in which six different prizes can be distributed among three children each receiving at least one prize is-

The correct option is B 540Prizes can be distributed number of 1,2,3 or 2,2,2, or 1,4,1 to 3 children.Total no. of ways =6!3!2!×3!+6!2!2!2!3!×3!+6!4!2!×31 =360+90+90=540Hence, option (B) is correct.