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  4. A team of four students is to be selected from a total of 12 students. Total number of ways in which the team can be selected such that two particular students refuse to be together and other two particular students wish to be together only, is equal to

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Q. A team of four students is to be selected from a total of $12$ students. Total number of ways in which the team can be selected such that two particular students refuse to be together and other two particular students wish to be together only, is equal to

Permutations and Combinations

Solution:

Let $S_{1}$ and $S_{2}$ refuse to be together and $S_{3}$ and $S_{4}$ want to be together only. Totall ways when $S_{3}$ and $S_{4}$ are selected
$=\left({ }^{8} C_{2}+{ }^{2} C_{1} \times{ }^{8} C_{1}\right)=44$
Total ways when $S_{3}$ and $S_{4}$ are not selected
$=\left({ }^{8} C_{4}+{ }^{2} C_{1} \times{ }^{8} C_{3}\right)=182$
Thus, total ways $=44+182=226$