If order matters, then there are 12 ways to do this
If order does not matter, then there are 6 ways to do this
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We have 4 choices for the first slot and 3 choices for the next (we can't reuse a letter) so that's where 4*3 = 12 comes from
If order doesn't matter, then something like AB is the same as BA. So we are doubly counting each possible combo. To fix this, we divide by 2: 12/2 = 6
To be more formal, you can use nPr and nCr to get 12 and 6 respectively (use n = 4 and r = 2)
Answer:
A. a + c ≥ 70; 15a + 4c ≤ 1000
Step-by-step explanation:
<u>Given data:</u>
A community action committee is planning a block party
The party will be successful if at least 70 adults (a) and children (c) attend
Their budget for the party is $1,000
The cost for each adult will be $15
And the cost for each child will be $4
<u>Solution:</u>
<u />
At least 70 adults and children attend.
So the inequality here is a + c ≥ 70
The budget for the party is $1,000.
So the cost for all adults plus the cost for all children cannot exceed $1,000
And hence the inequality here is; 15a + 4c ≤ 1,000
I think it’s the third option
