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Question Video: Using Properties of Combinations to Find an Unknown Mathematics • Second Year of Secondary School

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Find π‘Ÿ given that 105πΆπ‘Ÿ > 1 and π‘ŸπΆ103 > 1.

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Video Transcript

Find π‘Ÿ given that 105 πΆπ‘Ÿ is greater than one and π‘ŸπΆ 103 is greater than one.

To answer this question, we recall the definition of a combination. The combination π‘›πΆπ‘˜ represents the number of different ways to select π‘˜ objects from a total of 𝑛 distinct objects, where the order does not matter.

Given that 𝑛 and π‘˜ are positive integers and 𝑛 is greater than or equal to π‘˜, the formula for a combination is π‘›πΆπ‘˜ equals 𝑛 factorial divided by 𝑛 minus π‘˜ factorial multiplied by π‘˜ factorial. We are asked to find the value of π‘Ÿ, satisfying both 105 πΆπ‘Ÿ is greater than one and π‘ŸπΆ 103 is greater than one.

Let’s consider what we know about π‘Ÿ from the combination 105 πΆπ‘Ÿ. In this case, 𝑛 equals 105 and π‘˜ equals π‘Ÿ. We know in general that π‘˜ is greater than zero and 𝑛 is greater than or equal to π‘˜. Therefore, in this case, π‘Ÿ is greater than zero and 105 is greater than or equal to π‘Ÿ.

Now we will consider the combination π‘ŸπΆ 103. Here, we have 𝑛 equals π‘Ÿ and π‘˜ equals 103. According to the definition of a combination, in this case, π‘Ÿ is greater than zero and π‘Ÿ is greater than or equal to 103. We already know that π‘Ÿ is a positive integer. However, we now have more specific information about the value of π‘Ÿ. We have 105 is greater than or equal to π‘Ÿ, and we have π‘Ÿ is greater than or equal to 103. Pulling these statements together, we have π‘Ÿ is greater than or equal to 103 and less than or equal to 105.

Let’s consider what positive integers are in this range. The possible values for π‘Ÿ include 103, 104, and 105. Let’s see what happens with π‘ŸπΆ 103 if π‘Ÿ equals 103. If π‘Ÿ equals 103, then we have 103 𝐢 103, which equals 103 factorial divided by 103 minus 103 factorial multiplied by 103 factorial. 103 minus 103 factorial simplifies to zero factorial which equals one. After simplifying the denominator, we have 103 factorial divided by 103 factorial. Thus, 103 𝐢 103 equals one.

But this is a problem because we are told that π‘ŸπΆ 103 must be greater than one. Therefore, π‘Ÿ cannot equal 103. Something similar happens with 105 πΆπ‘Ÿ, when we let π‘Ÿ equal 105. We find that 105 𝐢 105 equals one as well. Therefore, π‘Ÿ cannot equal 105. In fact, any combination of the form π‘ŸCπ‘Ÿ will always equal one, because there is only one way to select all the objects.

It seems that 104 is the only possibility that remains. To be sure 104 satisfies both given inequalities, we will substitute 104 for π‘Ÿ. If π‘Ÿ equals 104 and we use the first combination formula 105 πΆπ‘Ÿ, we find that 105 𝐢 104 equals 105, and 105 is certainly greater than one.

Now we move on, to check π‘Ÿ equals 104 in the second combination π‘ŸπΆ 103. We find that 104 𝐢 103 equals 104. This satisfies the second given inequality, because 104 is greater than one.

In conclusion, we have shown that π‘Ÿ equals 104, which satisfies both of the conditions given.

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