You visited us 81 times! Enjoying our articles? Unlock Full Access!

A+B+C=N (N Varying)

In how many w...

Question

In how many ways you can distribute at most 10 apples among A, B, C such that A and B should get at least 1 and 2 apples respectively?

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

286

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

120

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

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

Open in App

Solution

The correct option is C 120
We have a + b + c $\leq$ 10 and a $\geq$ 1 and b $\geq$ 2
Now, a + b + c + d = 10
Giving one to A and two apples to B, we get
a + b + c + d = 7
Now total number of ways = $^{(7 + 4 - 1)} C_{(4 - 1)} =^{10} C_{3} = 120$ .

Suggest Corrections

Similar questions

Q. In how many ways

12

apples can be distributed among

4

children such that every child gets at least

2

apples?

Q. Find the number of ways in which

8

non-identical apples can be distributed among

3

boys such that every boy should get at least

1

apple & at most

4

apples.

How many ways can Adarsh distribute 17 identical apples among 6 students if he can also keep some apples for himself and each student should get at least one apple and not more than 6?

A basket has 5 oranges and 4 apples. In how many ways can you make a selection if you have to take at least 1 orange and 1 apple?

Q. In how many ways Ram can distributed

40

identical apples to his six children named

A, B, C, D, E

and

F

such that

A

gets two more than

B, C

gets

3

more than

F

and

D

gets five less then

E

and any one gets at least on fruit

Join BYJU'S Learning Program

In how many ways you can distribute at most 10 apples among A, B, C such that A and B should get at least 1 and 2 apples respectively?

The correct option is C 120We have a + b + c ≤ 10 and a ≥ 1 and b ≥ 2 Now, a + b + c + d = 10 Giving one to A and two apples to B, we get a + b + c + d = 7 Now total number of ways = (7+4−1)C(4−1)=10C3=120.

The correct option is C 120
We have a + b + c $\leq$ 10 and a $\geq$ 1 and b $\geq$ 2
Now, a + b + c + d = 10
Giving one to A and two apples to B, we get
a + b + c + d = 7
Now total number of ways = $^{(7 + 4 - 1)} C_{(4 - 1)} =^{10} C_{3} = 120$ .