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Doss [256]
3 years ago
6

A team of 17 softball players need to choose three players to refill the water cooler. How many different ways are there to sele

ct 3 players?
Mathematics
1 answer:
KATRIN_1 [288]3 years ago
5 0
<h3>Answer: 680 different combinations</h3>

=======================================================

Explanation:

If order mattered, then we'd have 17*16*15 = 4080 different permutations. Notice how I started with 17 and counted down 1 at a a time until I had 3 slots to fill. We count down by 1 because each time we pick someone, we can't pick them again.

So we have 4080 different ways to pick 3 people if order mattered. But again order doesn't matter. All that counts is the group itself rather than the individual or how they rank. There are 3*2*1 = 6 ways to order any group of three people, which means there are 4080/6 = 680 different combinations possible.

An alternative is to use the nCr formula with n = 17 and r = 3. That formula is

_n C _r = \frac{n!}{r!*(n-r)!}

where the exclamation marks indicate factorials

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3 years ago
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2 years ago
Which of the following is the expansion of (3c + d2)6?
vovangra [49]

Answer: Option A) 729c^6 + 1,458c^5d^2 + 1,215c^4d^4 + 540c^3d^6 + 135c^2d^8 + 18cd^{10} + d^{12} is the correct expansion.

Explanation:

on applying binomial theorem,  (a+b)^n=\sum_{r=0}^{n} \frac{n!}{r!(n-r)!} a^{n-r} b^r

Here a=3c, b=d^2 and n=6,

Thus, (3c+d^2)^6=\sum_{r=0}^{6} \frac{6!}{r!(6-r)!} (3c)^{n-r} (d^2)^r

⇒ (3c+d^2)^6= \frac{6!}{(6-0)!0!} (3c)^{6-0}.(d^2)^0+\frac{6!}{(6-1)!1!} (3c)^{6-1}.(d^2)^1+\frac{6!}{(6-2)!2!} (3c)^{6-2}.(d^2)^2+\frac{6!}{(6-3)!3!} (3c)^{6-3}.(d^2)^3+\frac{6!}{(6-4)!4!} (3c)^{6-4}.(d^2)^4+\frac{6!}{(6-5)!5!} (3c)^{6-5}.(d^2)^5+\frac{6!}{(6-6)!6!} (3c)^{6-6}.(d^2)^6

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⇒(3c+d^2)^6=(3c)^6.d^0+\frac{720}{120} (3c)^5.d^2+\frac{720}{48} (3c)^4.d^4+\frac{720}{36} (3c)^3.d^6+\frac{720}{48} (3c)^2.d^8+\frac{720}{120} (3c).d^{10}+.d^{12}

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