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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?

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286

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120

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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 = $^{(7 + 4 - 1)} C_{(4 - 1)} =^{10} C_{3} = 120$ .

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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 = $^{(7 + 4 - 1)} C_{(4 - 1)} =^{10} C_{3} = 120$ .