Thirty-six people in student council are running for the offices of president and vice-president. In how many different ways can those offices be assigned? 1,260 578 71
36 P 2
\[\large{^{36}C_{2} \times 2!}\]
C - combination P - permutation
ok so how do i do a combo
\[\large{^n C_{r} = \cfrac{n!}{r!(n-r)!}}\]
\[\large{^n P_r = \cfrac{n!}{(n-r)!}}\]
Actually: \[\large{^{36} P_{2} = ^{36}C_{2} \times 2!}\]
ok thank you
:) Your welcome
hmm
C - choosing P - arranging (another way of understanding) You can either directly arrange 2 people out of 36 - 36P2 OR Choose 2 out of 36 and then arrange them - 36C2 * 2!
so would it be 578?
36 choices for president one that is chosen 35 for veep by the counting principle the number of ways to do this is \(36\times 35\) do not get married to any formulas
1260
Both are same @satellite73
all these formulas are based on formalizing the counting principle for different scenarios think don't be a slave to the formulas
Ok, that makes a bit more scense
\[\large{^{36} C_{2} \times 2! = \cfrac{36!}{34!} = 36*35 = 1260}\]
I agree with @satellite73 . Don't use formulas unless necessary. When you can solve the problems without them then go for it
yes, i know both are the same but it is clear that 36 choices followed by 36 choices gives \(36\times 35\) choices based on nothing more than common sense and the counting principle no formulas needed no factorials nothing of the kind
typo there but you get the idea
Correct ^^^ Special cases/formulas are useless if you have the basic concepts correct
Though remember this: @satellite73 used `36 * 35` because it is an AND operation. You have to choose president AND vice- president
vishweshshrimali5 , why do write \(\normalsize\color{blue}{ ~^{\LARGE36}P_{\LARGE2} }\) instead of \(\normalsize\color{blue}{ ~_{\LARGE36}P_{\LARGE2} }\) ?
Similarly, I too used `36C2 * 2!` because I have to choose 2 people out of 36 AND then I have to arrange them
That is notation I have always read and studied :)
further, even \(\binom{n}{k}\) should not be formula driven imagine computing \[\binom{100}{2}\] when all that is required is \[\frac{100\times 99}{2}\]
Yep :)
thanks you guys, you both have been a great help
Things are easier to understand and solve when you know the "WHY"
Your welcome @CLoudrunner :)
@SolomonZelman i have never seen \(_nC_k\) or \(^nC_k\) in a text only \(\binom{n}{k}\)
I have seen both \(\large{^nC_k}\) and \(\large{\binom{n}{k}}\) - this one in foreign writers' books
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