Answer :
Let's solve the given problem step by step:
The problem states that:
[tex]\[ \frac{{}^nC_6}{{}^{n-2}C_3} = \frac{91}{4} \][/tex]
We need to determine the value of [tex]\( n \)[/tex].
### Step 1: Understanding Combinations
The combination formula is given by:
[tex]\[ {}^nC_k = \frac{n!}{k!(n-k)!} \][/tex]
So, let's first write out the combinations:
[tex]\[ {}^nC_6 = \frac{n!}{6!(n-6)!} \][/tex]
[tex]\[ {}^{n-2}C_3 = \frac{(n-2)!}{3!(n-5)!} \][/tex]
### Step 2: Set Up the Given Equation
We substitute these combinations into the given equation:
[tex]\[ \frac{\frac{n!}{6!(n-6)!}}{\frac{(n-2)!}{3!(n-5)!}} = \frac{91}{4} \][/tex]
### Step 3: Simplify the Fraction
To simplify the left-hand side of the equation:
[tex]\[ \frac{\frac{n!}{6!(n-6)!}}{\frac{(n-2)!}{3!(n-5)!}} = \frac{n! \cdot 3!(n-5)!}{6!(n-6)! \cdot (n-2)!} \][/tex]
Notice that [tex]\((n-2)! = (n-2)(n-3)(n-4)(n-5)!\)[/tex], so we can also cancel [tex]\((n-5)!\)[/tex] from numerator and denominator:
[tex]\[ \frac{n! \cdot 3!}{6! \cdot (n-2)(n-3)(n-4)} \][/tex]
Since [tex]\( 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 \)[/tex]:
[tex]\[ \frac{n(n-1)(n-2)(n-3)(n-4)(n-5) \cdot 6}{720 \cdot (n-2)(n-3)(n-4)} = \frac{91}{4} \][/tex]
### Step 4: Cancel Common Terms
Now we can cancel [tex]\((n-2)(n-3)(n-4)\)[/tex]:
[tex]\[ \frac{n(n-1)(n-5) \cdot 6}{120} = \frac{91}{4} \][/tex]
Simplifying [tex]\( 120 \)[/tex] as [tex]\( 6 \times 20 \)[/tex]:
[tex]\[ \frac{n(n-1)(n-5)}{20} = \frac{91}{4} \][/tex]
### Step 5: Solve the Equation
Multiplying both sides by 20 to eliminate the denominator:
[tex]\[ n(n-1)(n-5) = \frac{91 \times 20}{4} \][/tex]
[tex]\[ n(n-1)(n-5) = 91 \times 5 \][/tex]
[tex]\[ n(n-1)(n-5) = 455 \][/tex]
### Step 6: Find the Value of [tex]\( n \)[/tex]
To solve [tex]\( n(n-1)(n-5) = 455 \)[/tex], we need to test plausible values of [tex]\( n \)[/tex].
Testing some values sequentially (starting above [tex]\( n = 6\)[/tex]) would show whether they satisfy the equation.
Upon trying different [tex]\( n \)[/tex], it becomes evident that no integer value [tex]\( n \)[/tex] within a reasonable range satisfies the equation based on the conditions stated:
Thus,
[tex]\[ \boxed{\text{Solution not found within range.}} \][/tex]
The problem states that:
[tex]\[ \frac{{}^nC_6}{{}^{n-2}C_3} = \frac{91}{4} \][/tex]
We need to determine the value of [tex]\( n \)[/tex].
### Step 1: Understanding Combinations
The combination formula is given by:
[tex]\[ {}^nC_k = \frac{n!}{k!(n-k)!} \][/tex]
So, let's first write out the combinations:
[tex]\[ {}^nC_6 = \frac{n!}{6!(n-6)!} \][/tex]
[tex]\[ {}^{n-2}C_3 = \frac{(n-2)!}{3!(n-5)!} \][/tex]
### Step 2: Set Up the Given Equation
We substitute these combinations into the given equation:
[tex]\[ \frac{\frac{n!}{6!(n-6)!}}{\frac{(n-2)!}{3!(n-5)!}} = \frac{91}{4} \][/tex]
### Step 3: Simplify the Fraction
To simplify the left-hand side of the equation:
[tex]\[ \frac{\frac{n!}{6!(n-6)!}}{\frac{(n-2)!}{3!(n-5)!}} = \frac{n! \cdot 3!(n-5)!}{6!(n-6)! \cdot (n-2)!} \][/tex]
Notice that [tex]\((n-2)! = (n-2)(n-3)(n-4)(n-5)!\)[/tex], so we can also cancel [tex]\((n-5)!\)[/tex] from numerator and denominator:
[tex]\[ \frac{n! \cdot 3!}{6! \cdot (n-2)(n-3)(n-4)} \][/tex]
Since [tex]\( 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 \)[/tex]:
[tex]\[ \frac{n(n-1)(n-2)(n-3)(n-4)(n-5) \cdot 6}{720 \cdot (n-2)(n-3)(n-4)} = \frac{91}{4} \][/tex]
### Step 4: Cancel Common Terms
Now we can cancel [tex]\((n-2)(n-3)(n-4)\)[/tex]:
[tex]\[ \frac{n(n-1)(n-5) \cdot 6}{120} = \frac{91}{4} \][/tex]
Simplifying [tex]\( 120 \)[/tex] as [tex]\( 6 \times 20 \)[/tex]:
[tex]\[ \frac{n(n-1)(n-5)}{20} = \frac{91}{4} \][/tex]
### Step 5: Solve the Equation
Multiplying both sides by 20 to eliminate the denominator:
[tex]\[ n(n-1)(n-5) = \frac{91 \times 20}{4} \][/tex]
[tex]\[ n(n-1)(n-5) = 91 \times 5 \][/tex]
[tex]\[ n(n-1)(n-5) = 455 \][/tex]
### Step 6: Find the Value of [tex]\( n \)[/tex]
To solve [tex]\( n(n-1)(n-5) = 455 \)[/tex], we need to test plausible values of [tex]\( n \)[/tex].
Testing some values sequentially (starting above [tex]\( n = 6\)[/tex]) would show whether they satisfy the equation.
Upon trying different [tex]\( n \)[/tex], it becomes evident that no integer value [tex]\( n \)[/tex] within a reasonable range satisfies the equation based on the conditions stated:
Thus,
[tex]\[ \boxed{\text{Solution not found within range.}} \][/tex]