Aptitude Discussion

**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:

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

✖ A. |
$3! × {^5C_3} × {^{15}C_3}$ |

✖ B. |
$3! × {^5P_3} × {^{15}P_3}$ |

✔ C. |
$6! × {^5C_3} × {^{15}C_3}$ |

✖ D. |
$6! × {^5P_3} × {^{15}P_3}$ |

**Solution:**

Option(**C**) 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 is

$6! × {^5C_3} × {^{15}C_3}$

**Vejayanantham TR**

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**Lee**

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If we know that the 3 freshmen won the first three prizes, and the next three students got the next three prizes then shouldnt we consider the respective permutations separately? like 3!*3!

**Arpit SAxena**

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Please someone expain the question to me.

The solution seems rather confusing.

3 fishermen in 3 prize at any order = combination = 5C3

Remaining, 3 students from 15 in any order = combination = 15C3

Now, we need to find the order => total 6 prizes for 6 students => Order = combination = 6!

Ans = 5C3*15C3*6!