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Math  /  Algebra

QuestionQuestion 26: If nP3=4nC5{ }^{n} P_{3}=4^{n} C_{5}, then nn equals
Single-digit integer (-9 to 9)
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Studdy Solution

STEP 1

1. nP3 {}^{n}P_{3} represents the number of permutations of 3 items from a set of n n items.
2. nC5 {}^{n}C_{5} represents the number of combinations of 5 items from a set of n n items.
3. The equation nP3=4×nC5 {}^{n}P_{3} = 4 \times {}^{n}C_{5} needs to be solved for n n .

STEP 2

1. Express the permutation and combination formulas.
2. Substitute these expressions into the given equation.
3. Simplify the equation and solve for n n .

STEP 3

Write the formula for permutations and combinations:
The formula for permutations is: nPr=n!(n−r)! {}^{n}P_{r} = \frac{n!}{(n-r)!}
The formula for combinations is: nCr=n!r!(n−r)! {}^{n}C_{r} = \frac{n!}{r!(n-r)!}
For this problem, we have: nP3=n!(n−3)! {}^{n}P_{3} = \frac{n!}{(n-3)!} nC5=n!5!(n−5)! {}^{n}C_{5} = \frac{n!}{5!(n-5)!}

STEP 4

Substitute these expressions into the given equation:
n!(n−3)!=4×n!5!(n−5)! \frac{n!}{(n-3)!} = 4 \times \frac{n!}{5!(n-5)!}

STEP 5

Cancel n! n! from both sides of the equation:
1(n−3)!=45!(n−5)! \frac{1}{(n-3)!} = \frac{4}{5!(n-5)!}

STEP 6

Cross-multiply to eliminate the fractions:
5!(n−5)!=4(n−3)! 5!(n-5)! = 4(n-3)!

STEP 7

Recognize that (n−3)!=(n−3)(n−4)(n−5)! (n-3)! = (n-3)(n-4)(n-5)! . Substitute this into the equation:
5!(n−5)!=4(n−3)(n−4)(n−5)! 5!(n-5)! = 4(n-3)(n-4)(n-5)!
Cancel (n−5)! (n-5)! from both sides:
5!=4(n−3)(n−4) 5! = 4(n-3)(n-4)

STEP 8

Calculate 5! 5! and solve for n n :
5!=120 5! = 120
120=4(n−3)(n−4) 120 = 4(n-3)(n-4)
Divide both sides by 4:
30=(n−3)(n−4) 30 = (n-3)(n-4)
Expand the right side:
30=n2−7n+12 30 = n^2 - 7n + 12
Rearrange to form a quadratic equation:
n2−7n+12−30=0 n^2 - 7n + 12 - 30 = 0
n2−7n−18=0 n^2 - 7n - 18 = 0

STEP 9

Factor the quadratic equation:
(n−9)(n+2)=0 (n - 9)(n + 2) = 0
Set each factor to zero and solve for n n :
n−9=0⇒n=9 n - 9 = 0 \quad \Rightarrow \quad n = 9
n+2=0⇒n=−2 n + 2 = 0 \quad \Rightarrow \quad n = -2
Since n n must be a positive integer (as it represents a count of items), we choose:
n=9 n = 9
The value of n n is:
9 \boxed{9}

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