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If there are ten positive real numbers n1 < n2 < n3 … < n10, how many

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If there are ten positive real numbers n1 < n2 < n3 … < n10, how many [#permalink]

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New post 11 Nov 2010, 04:02
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If there are ten positive real numbers n1 < n2 < n3 … < n10, how many triplets of these numbers (n1, n2, n3), (n2, n3, n4) … can be generated such that in each triplet the first number is always less than the second number, and the second number is always less than the third number?

a) 45
b) 90
c) 120
d) 180
e) 150

One solving method is the following:
Three numbers can be selected and arranged out of ten numbers in 10P3 ways=10!/7!=10*9*8. Now this arrangement is restricted to a given condition that first number is always less than the second number, and the second number is always less than the third number. Hence three numbers can be arranged among themselves in 3! ways.

Required number of arrangements=(10*9*8)/(3*2)=120

(the source: Winners’ Guide to GMAT Math – Part II)

Pls, can someone explain me, why 10P3 is divided by 3! ? Inasmuch as I understand the denominator denotes that arragements cases of repeated numbers are to be excluded. Take we this into account then 3! in the denominator has to do with cases such as (n1, n1, n1) or (n2, n2, n2) … But, the question asked in this problem is of a different kind. Then what does 3! mean or has 3! actully to do with triplets with repeated numbers which I could not comprehend?

Many thanks beforhand for detailed explanations !!!
[Reveal] Spoiler: OA
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Re: If there are ten positive real numbers n1 < n2 < n3 … < n10, how many [#permalink]

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New post 11 Nov 2010, 04:15
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feruz77 wrote:
If there are ten positive real numbers n1 < n2 < n3 … < n10, how many triplets of these numbers (n1, n2, n3), (n2, n3, n4) … can be generated such that in each triplet the first number is always less than the second number, and the second number is always less than the third number?

a) 45
b) 90
c) 120
d) 180
e) 150

One solving method is the following:
Three numbers can be selected and arranged out of ten numbers in 10P3 ways=10!/7!=10*9*8. Now this arrangement is restricted to a given condition that first number is always less than the second number, and the second number is always less than the third number. Hence three numbers can be arranged among themselves in 3! ways.

Required number of arrangements=(10*9*8)/(3*2)=120

(the source: Winners’ Guide to GMAT Math – Part II)

Pls, can someone explain me, why 10P3 is divided by 3! ? Inasmuch as I understand the denominator denotes that arragements cases of repeated numbers are to be excluded. Take we this into account then 3! in the denominator has to do with cases such as (n1, n1, n1) or (n2, n2, n2) … But, the question asked in this problem is of a different kind. Then what does 3! mean or has 3! actully to do with triplets with repeated numbers which I could not comprehend?

Many thanks beforhand for detailed explanations !!!


Consider the following approach: we can choose \(C^3_{10}=120\) different triplets out of 10 distinct numbers. Each triplet (for example: {a,,b,c}) can be arranged in 3! ways ({a,b,c}; {a,c,b}, {b,c,a}, ...), but only one arrangement (namely {a,b,c}) will be in ascending order, so basically \(C^3_{10}=120\) directly gives the desired # of such triplets.

If you look at the approach you posted: \(\frac{P^3_{10}}{3!}\) then you can notice that it's basically the same because: \(\frac{P^3_{10}}{3!}=C^3_{10}=120\)

Answer: C.

Similar problems:
probability-of-picking-numbers-in-ascending-order-89035.html?hilit=ascending%20probability
probability-for-consecutive-numbers-102191.html?hilit=ascending%20probability#p793614

Hope it's clear.
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Re: If there are ten positive real numbers n1 < n2 < n3 … < n10, how many [#permalink]

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Re: If there are ten positive real numbers n1 < n2 < n3 … < n10, how many   [#permalink] 29 Aug 2016, 09:29
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