Mathematics · Permutations and Combinations

JEE Main 2024 — 5 April, Shift 1 — Question 24

The number of ways of getting a sum 16 on throwing a dice four times is \qquad

Answer: 125

Numerical answer — enter this value.

Step-by-step solution

(x1+x2….+x6)4\left(x^{1}+x^{2} \ldots .+x^{6}\right)^{4}

x4⋅(1−x61−x)4\mathrm{x}^{4} \cdot\left(\frac{1-\mathrm{x}^{6}}{1-\mathrm{x}}\right)^{4}

x4⋅(1−x6)4⋅(1−x)−4x^{4} \cdot\left(1-x^{6}\right)^{4} \cdot(1-x)^{-4}

x4[1−4x6+6x12….][(1−x)−4]\mathrm{x}^{4}\left[1-4 \mathrm{x}^{6}+6 \mathrm{x}^{12} \ldots.\right]\left[(1-\mathrm{x})^{-4}\right]

(x4−4x10+6x16….)(1−x)−4\left(x^{4}-4 x^{10}+6 x^{16} \ldots.\right)(1-x)^{-4}

(x4−4x10+6x16)(1+15C12x12+9C6x6….)\left(x^{4}-4 x^{10}+6 x^{16}\right)\left(1+{ }^{15} C_{12} x^{12}+{ }^{9} C_{6} x^{6} \ldots.\right) (15C12−4⋅9C6+6)x16\left({ }^{15} \mathrm{C}_{12}-4 \cdot{ }^{9} \mathrm{C}_{6}+6\right) \mathrm{x}^{16}

(15C3−4⋅9C6+6)\left({ }^{15} \mathrm{C}_{3}-4 \cdot{ }^{9} \mathrm{C}_{6}+6\right)

=35×13−6×8×7+6=35 \times 13-6 \times 8 \times 7+6

=455−336+6=455-336+6

=125=125

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Subject
Mathematics
Chapter
Permutations and Combinations
Topic
Number of integral solution of linear Equations
The number of ways of getting a sum 16 on throwing a dice four times… | JEE Main 2024 PYQ with Solution · DhiX AI