Question

if nc4 =5×np3 , find the value of n​

Answer 1

Step-by-step explanation:

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Answer 2

The value of n is 123.

• Combinations are ways to choose elements from a collection in mathematics where the order of the selection is irrelevant. Let's say we have a trio of numbers: P, Q, and R. Then, combination determines how many ways we can choose two numbers from each group.
• It is easy to count the number of combinations in smaller cases, but the likelihood of a set of combinations is also higher in cases with many groups of components or sets. The definition of the combination is "An arrangement of objects where the order of the objects is irrelevant." Combining these two words indicates "Selection of things," where the order of the items is irrelevant.

Here, according to the given information, we are given that,

$nC_{4} =5.nP_{3}$

Now, $nC_{4} =\frac{n!}{4!.(n-4)!}$ and $nP_{3} =\frac{n!}{(n-3)!}$

Then, we have,

$nC_{4} =5.nP_{3}$

Or, $\frac{n!}{4!.(n-4)!} = 5.\frac{n!}{(n-3)!}$

Or, $\frac{(n-3)(n-4)!}{4!.(n-4)!} =5.$

Or, $n-3 = 5.4.3.2.1$

Or, $n-3 = 120.$

Or, $n=123.$

Hence, the value of n is 123.

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