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Class 11MathematicsPermutations and CombinationsCombinationsExpert
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For (x_1+x_2+x_3+x_4=20), with (x_1\ge5) and (x_2,x_3,x_4\ge1), how many positive integer solutions are there?

Answer and explanation

Correct answer: (455)

After subtracting minimums (5,1,1,1), (12) remains, so ( \binom{15}{3}=455 ). Subtract unequal minimum conditions first.

Related tags

Class11CombinationsPositive-SolutionsStars-And-BarsExpertPermutations And CombinationsMathematicsClass 11 Mcq
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