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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    $\begingroup$ You also need the multiplication principle. The student selects three from the first set, then three from the second, then three from the third. $\endgroup$ – David Mitra Jul 1 '13 at 16:47
  • $\begingroup$ Can you please elaborate $\endgroup$ – SSK Jul 1 '13 at 16:48
  • $\begingroup$ 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. $\endgroup$ – David Mitra Jul 1 '13 at 16:50

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:

$$\binom 53 \cdot \binom 53 \cdot \binom 53 \quad \text{ways of doing so}$$

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  • $\begingroup$ do you know the binomial coefficient? $\binom 53 \neq \frac 53$ $$\binom 53 = \frac{5!}{3!2!} = \frac{5\cdot 4}{2\cdot 1} = 10.$$ $\endgroup$ – amWhy Jul 1 '13 at 17:07
  • $\begingroup$ Oh its combination formula .. thanks Understood now ! $\endgroup$ – SSK Jul 1 '13 at 17:09
  • $\begingroup$ You're welcome, yes: it's the notation used to denote $5$-choose-$3$. $\endgroup$ – amWhy Jul 1 '13 at 17:10
  • $\begingroup$ @amWhy: I even had a grammar error in the comment I made to you! Happens all the time - I sometimes edit twn times! ;-) $\endgroup$ – Amzoti Jul 2 '13 at 2:52

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 $={.\choose.}{.\choose.}{.\choose.}$

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  • $\begingroup$ $9*6*3$ which is wrong answer $\endgroup$ – SSK Jul 1 '13 at 16:52
  • $\begingroup$ Number of ways of selecting $3$ objects out of $5={5\choose 3}$ $\endgroup$ – Aang Jul 1 '13 at 17:10

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