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Common Information

There are $5$ freshmen, $8$ sophomores, and $7$ juniors in a chess club. Find the number of orders in which the $6$ students from this club can win the first six prizes:


Common Information Question: 2/4

If all students attend the competition and the winners are exactly 3 freshmen and 3 sophomores.


$3! × {^5C_3} × {^15C_3}$


$3! × {^5C_3} × {^8C_3}$


$6! × {^5P_3} × {^15P_3}$


$6! × {^5C_3} × {^8C_3}$

 Hide Ans

Option(D) is correct

We need to think in terms of permutations rather than combinations.

Can we simply use the same procedure and change combinations to permutations?

For example, is the number of orders in which 3 freshmen and 3 juniors or seniors win prizes equal to ${^5P_3} × {^{15}P_3}$?

This would give us the number of possible orders only if the three freshmen win the first three prizes in some order followed by the three sophomores or juniors winning the next 3 prizes in some order.

We need to consider the number of possible orders for all six students who are chosen.

Recall that: $\dfrac{^nP_r}{r!} = {^nC_r}$
Therefore, ${^nP_r} = r! × {^nC_r}$

We can find the number of orders in which the 6 students win prizes by multiplying the combinations by 6!

The number of orders of 6 students with exactly 3 freshmen and 3 sophomores is:
$6! × {^5C_3} × {^8C_3}$

(2) Comment(s)


thank you deepak for explaining this question.


Say, Oranges $= o$, Apples $= a$ and Banana $= b$

From the give information we can form two equations:

$2(o)+3a+5b=40$ -------- (i)

$3(o)+5a+9b=54$ -------- (ii)

Finally we to find the price of 1 orange($o$), 1 apple($a$) and 1 banana($b$).

We would be needing 3 equations to find the values of $o, a$ and $b$. but we have only 2 equations Sad

But here is the catch Wink

Upon minor examination we can manipulate the given to equations to get the 3rd required equation without solving the individual variables ($o, a$ and $b$).

So if we multiply the $1^{st}$ eq. by $2$ and subtract the $2^{nd}$ equation from it, we get what is asked in the question. This is written in the equation form as:

$2*(i) - (ii)$

$= [4(o)+6a+10b-80]- [3(o)+5a+9b-64]=0$

$= o+a+b-16=0$

Which gives the price of 1 orange, 1 apple and 1 banana as

$o+a+b= \textbf{ 16 Rs.}$ Smile