The number of different boat parties of 8, consisting of 5 boys and 3 girls, that can be, made from 10 boys and 10 girls, are

15120

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

60480

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

30240

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

none of these

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

Open in App

Solution

The correct option is A 15120
Total boys $= 25$
Total girls $= 10$
We have to select $5$ boys and $3$ girls
Then we have $=^{25} C_{5} \times^{10} C_{3}$
$= \frac{25!}{5! \times 20!} \times \frac{10!}{3! 7!}$
$= 6375600$ .

Suggest Corrections

Similar questions

How many different boat parties of 8, consisting of 8, consisting of 5 boys and 3 girls, can be made from 25 boys and 10 girls ?

Q. Consider a class of 5 girls and 7 boys. The number of different teams consisting of 2 girls and 3 boys that can be formed from this class, if there are two specific boys A and B, who refuse to be a member of same team, is :

A committee consists of 5 students 3 girls and 2 boys. If the team is to be formed out of 7 boys and 5 girls, then how many ways this selection can be made?

Q. A committee of

4

persons has to be chosen from

8

boys and

6

girls, consisting of at least one girl. Find the probability that the committee consists of more girls than boys.

Join BYJU'S Learning Program

The number of different boat parties of 8, consisting of 5 boys and 3 girls, that can be, made from 10 boys and 10 girls, are

The correct option is A 15120Total boys=25Total girls=10We have to select 5 boys and 3 girlsThen we have =25C5×10C3=25!5!×20!×10!3!7!=6375600.

The correct option is A 15120
Total boys $= 25$
Total girls $= 10$
We have to select $5$ boys and $3$ girls
Then we have $=^{25} C_{5} \times^{10} C_{3}$
$= \frac{25!}{5! \times 20!} \times \frac{10!}{3! 7!}$
$= 6375600$ .