A history examination is made up of $3$ set of $5$ Question each and a student must select $3$ questions from each set . How many different sets of $9$ questions can the student select ?
I am in dilemma how i can use combination formula
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2You also need the multiplication principle. The student selects three from the first set, then three from the second, then three from the third. – David Mitra Jul 01 '13 at 16:47
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Can you please elaborate – SSK Jul 01 '13 at 16:48
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How many ways can the student select three questions from the first set? How many ways for the second? The third? Multiply these numbers together. – David Mitra Jul 01 '13 at 16:50
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From each of the three sets of $5$ questions, a student must select three questions.
So, yes, we can compute three sets of combinations:
$(1)$ From the first set of $5$ questions, there are $\binom 53$ ways a student may select three questions.
$(2)$ Similarly, from the second set of $5$ questions, the student must select 3 questions to answer. Again, there are $\binom 53$ ways of doing so.
$(3)$ And likewise for the third set.
$(*)$ Then, as David Mitra suggested, we need the multiplication principle to get, overall, the number of ways a student can answer questions on the test, for a total of:
$$\binom 53 \cdot \binom 53 \cdot \binom 53 \quad \text{ways of doing so}$$
amWhy
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do you know the binomial coefficient? $\binom 53 \neq \frac 53$ $$\binom 53 = \frac{5!}{3!2!} = \frac{5\cdot 4}{2\cdot 1} = 10.$$ – amWhy Jul 01 '13 at 17:07
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@amWhy: I even had a grammar error in the comment I made to you! Happens all the time - I sometimes edit twn times! ;-) – Amzoti Jul 02 '13 at 02:52
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HINT: Since students need to select total $9$ questions, and $3$ questions must be selected from each of $3$ sets, therefore, exactly $3$ questions from each set needs to be selected and total number of ways of doing so $={.\choose.}{.\choose.}{.\choose.}$
Aang
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