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# A candidate is required to answer $6$ out of $10$ questions which are divided into two groups, each containing $5$ questions. He is not permitted to attempt more than $4$ questions from either group. Find the number of different ways in which the candidate can choose six questions.

Last updated date: 29th Feb 2024
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Hint: First we have to define what the terms we need to solve the problem are.
Since the question is to find the number of ways, we are going to use permutation and combination methods which we study on our schools to approach the given questions to find the number of ways, since the number of permutations of r-objects can be found from among n-things is ${}^n{P_r}$ where p refers to the permutation. Also, similarly for combination we have r-things and among n-things are${}^n{C_r}$.

Complete step-by-step solution:
Let the question is to find the different number of ways hence we use only a combination method to approach the given question, since candidates are asked to answer $6$ out of $10$ questions which is divided into two groups, which means one part will contain five questions and equally.
Therefore, the number of ways of choosing six questions from ten questions is ten of the combination in six
This means ${}^{10}{C_6}$ (possible of $10$ questions from six needs to choose)
Hence by combination formula ${}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$as derived as ${}^{10}{C_6} = \dfrac{{10 \times 9 \times 8 \times 7}}{{1 \times 2 \times 3 \times 4}}$ (factorial of ten is divided by the factorial of six and resultant)
Hence solving this we get ${}^{10}{C_6} = \dfrac{{10 \times 9 \times 8 \times 7}}{{1 \times 2 \times 3 \times 4}} = 210$ (by using the multiplication and division method)
Hence the number of ways of choosing six questions from ten questions is ten of the combinations in six is two hundredth one.
Now we are going to solve the attempting more than four questions from each group and then minus from total we get the desired results;
Hence number of attempting more that four questions from each group is multiplied by two (two groups)
$2 \times {}^5{C_5} \times {}^5{C_1} = 2 \times 1 \times 5 = 10$ (Five combination of five is first group chosen so and then five combination one yields only one to choose)
Therefore, the required total number of different ways is $210 - 10 = 200$ (total number minus the restriction)

Note: We can also able to solve the above give question by ${}^5{C_3} \times {}^5{C_3} + {}^5{C_4} \times {}^5{C_2} + {}^5{C_2} \times {}^5{C_4}$ which is the total number of ways choosing the resultant (first three question from five chosen and same again and we can’t choose more than four questions so four or two combination occurs twice)
Hence ${}^5{C_3} \times {}^5{C_3} + {}^5{C_4} \times {}^5{C_2} + {}^5{C_2} \times {}^5{C_4} = 100 + 50 + 50 = 200$, the formula for combination is ${}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$ the difference is one r factorial will be multiplied in denominator on the combination