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In how many ways the letter of the word 'FAILURE' can be arranged with the condition that the four vowels are always together?
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In how many ways the letter of the word 'FAILURE' can be arranged with...
Solution:

Given: 2nC3 : nC2 = 52:3

To find the value of n, we need to simplify the given ratio and solve for n.

1. Simplifying the Ratios:
The ratio 2nC3 : nC2 can be simplified as follows:

2nC3 = (2n!)/((2n-3)! * 3!)
nC2 = (n!)/((n-2)! * 2!)

Dividing 2nC3 by nC2, we get:

(2n!)/((2n-3)! * 3!) / (n!)/((n-2)! * 2!)

Simplifying further:

[(2n! * (n-2)! * 2!)] / [(n! * (2n-3)! * 3!)]

2. Simplifying Factorials:
We can simplify the factorials in the numerator and denominator:

(2n * (2n-1) * (2n-2)! * (n-2)! * 2!) / (n * (n-1)! * (2n-3)! * 3!)

Cancelling out common terms:

(2n * (2n-1) * 2!) / (n * (n-1) * 3!)

3. Cross-Multiplying:
Given that the simplified ratio is equal to 52:3, we can set up the equation:

(2n * (2n-1) * 2!) / (n * (n-1) * 3!) = 52/3

Cross-multiplying:

(2n * (2n-1) * 2!) * 3! = (n * (n-1) * 3!) * 52

Simplifying:

2n * (2n-1) * 2 * 3 * 2 = n * (n-1) * 3 * 3 * 52

4. Simplifying Further:
Further simplifying the equation:

8n(2n-1) = 9n(n-1) * 52

Expanding and rearranging:

16n^2 - 8n = 468n^2 - 468n

Simplifying:

452n^2 - 460n = 0

Dividing both sides by 4n:

113n - 115 = 0

5. Solving for n:
Adding 115 to both sides:

113n = 115

Dividing both sides by 113:

n = 115/113

Therefore, the value of n is approximately 1.0177.

In conclusion, the value of n is approximately 1.0177 when the ratio 2nC3 : nC2 is equal to 52:3.
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In how many ways the letter of the word 'FAILURE' can be arranged with the condition that the four vowels are always together?
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