Sagot :
Answer:
ANSWERS:
\large\rm 1. \: b +51.b+5
• The sum of a number, b, and five.
• A number, b, added by five.
\large \rm 2. \: 2x - 62.2x−6
• Twice the number, x, subtracted by six.
• Six less than the product of two and a number, x.
\large \rm 3. \: 3x - 5 = 503.3x−5=50
• Thrice the number, x, diminished by five is equal to fifty.
• Five less than the product of three and a number, y, is fifty.
\large \rm 4. \: 2a \times 6=234.2a×6=23
• The product of twice a number, x, and 6 is twenty-three.
• Twice a number, a, multiplied by six is twenty-three.
\large \rm 5. \: 5b + 6 = 1005.5b+6=100
• Five times a number, b, added by six is one hundred.
• Six more than the product of five and a number, b, is one hundred.
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Answer:
✒️COMBINATIONS
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\large\underline{\mathbb{ANSWER}:}
ANSWER:
\qquad \LARGE \:\: \rm 5 \: Groups5Groups
*Please read and understand my solution. Don't just rely on my direct answer*
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\large\underline{\mathbb{SOLUTION}:}
SOLUTION:
Since the letters are grouped, the order doesn't matter. Solve for the number of combinations does 5 letters picked 4 at a time.
\begin{gathered} \begin{aligned} & \bold{Formula:} \\ & \quad \boxed{\rm _nC_r = \frac{n!}{r!(n-r)!}} \end{aligned} \end{gathered}
Formula:
n
C
r
=
r!(n−r)!
n!
\begin{gathered} \rm _5C_4 = \frac{5!}{4!(5-4)!} \\ \end{gathered}
5
C
4
=
4!(5−4)!
5!
\begin{gathered} \rm _5C_4 = \frac{5!}{4! \,1!} \\ \end{gathered}
5
C
4
=
4!1!
5!
\begin{gathered} \rm _5C_4 = \frac{5 \cdot \cancel{4!}}{\cancel{4!} } \\ \end{gathered}
5
C
4
=
4!
5⋅
4!
\rm _5C_4 = 5
5
C
4
=5
Therefore, there are 5 groups of 4 letters that can be made from the word "house".
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