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  • The number of ways to distribute 30 identical candies among four children <img alt="\mathrm{C}_{1}, \mathrm{C}_{2}, \mathrm{C}_{3}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B

The number of ways to distribute 30 identical candies among four children \mathrm{C}_{1}, \mathrm{C}_{2}, \mathrm{C}_{3} and \mathrm{C}_{4} so that \mathrm{C}_{2} receives atleast 4 and atmost 7 candies, \mathrm{C}_{3} receives atleast 2 and atmost 6 candies, is equal to:

Option: 1

205


Option: 2

615


Option: 3

510


Option: 4

430


Answers (1)

best_answer

\mathrm{4 \leq c_{2} \leq 7} \\

\mathrm{2 \leq c_{3} \leq 6}

Now for solution of  \mathrm{C_{1}+C_{2}+C_{3}+C_{4}=30} with given conditions, the number of solutions equal

\mathrm{\begin{aligned} &\text { coefficient of } x^{30} \text { in }\left(1+x+x^{2}+\cdots\right) \left(x^{4}+x^{5}+x^{6}+x^{7}\right)\\ \end{aligned}}\\

 \mathrm{\left(x^{2}+x^{3}+x^{4}+x^{5}+x^{6}\right) \left(1+x+x^{2}+\ldots\right)}

\mathrm{ = \text { coefficient of } x^{30} \text { in } x^{6}\left(1+x+x^{2}+\cdots\right)^{2} }

\mathrm{\left(1+x+x^{2}+x^{3}\right)\left(1+x+x^{2}+\ldots+x^{4}\right)}

\mathrm{=\text { coofficient of } x^{24} \text { in } \frac{1}{(1-x)^{2}} \cdot \frac{\left(x^{4}-1\right)}{(x-1)} \cdot \frac{\left(x^{5}-1\right)}{(x-1)}}

\mathrm{=\text { coefficient of } x^{24} \text { in }\left(x^{4}-1\right)\left(x^{5}-1\right)(1-x)^{-4}}

\mathrm{= coefficient\: of\: x^{24} \: in \left(x^{9}-x^{5}-x^{4}+1\right)(1+4 x \left.+{ }^{5} \mathrm{C}_{2} x^{2}+\ldots . . .\right) }

\mathrm{={ }^{18} C_{15}-{ }^{22} C_{19}-{ }^{23} C_{20}+{ }^{27} C_{24}} \\

\mathrm{={ }^{18} C_{3}-{ }^{22} C_{3}-{ }^{23} C_{3}+{ }^{27} C_{3}} \\

\mathrm{=816-1540-1771+2925} \\

\mathrm{=430}

Hence correct option is 4

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Gunjita

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