How to do 206th question?

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3 Answers
Mar 28, 2018

Answer: (#2#) #10!#

Explanation:

#10# men and #10# women to form #10# male-female couples.

Let the #10# men be fixed and the females be deciding how the couples will be matched.

The #1^(st)# woman will be able to choose between #10# man.

The #2^(nd)# woman will be able to choose between #9# man.

The #3^(rd)# woman will be able to choose between #8# man.

.
.
.
The #10^(th)# woman will be able to choose between #1# man.

Hence, the number of possible outcomes will be,

#10xx9xx8xx7xx6xx5xx4xx3xx2xx1=10!#

Therefore, there is #10!# number of ways to divide #10# man and #10# woman into #10# male-female couples.

Mar 30, 2018

See the response from my associate:

Explanation:

The question is slightly ambiguous in that it does not tell us whether the division into couples is ordered or not. In practice, the only option that works is 2) 10! with the understanding that the order of the couples does not matter. Given that, each man in turn could choose 1 woman in turn, so 10 choices for the first man 9
for the second, etc, giving 10! possible divisions into 10
couples.

Mar 30, 2018

Solution Part 2 of 2
Investigating value.

Logically you would expect a whole number solution and the only one giving that is option 2

Explanation:

#color(blue)("Which option do we select ?")#

#color(green)("Option 1") color(red)(larr" Not a whole number !")#

#color(white)()^20 C_10->(n!)/((n-r)!r!) ->(20!)/((20-10)!10!)=(20!)/((10!)^2)#

#(cancel(20)^2xx19xxcancel(18)^2xx17xxcancel(16)^2xxcancel(15)^3xxcancel(14)^2xx13xxcancel(12)^2xx11xxcancel(10!))/((cancel(10)^1xxcancel(9)^1xxcancel(8)^1xxcancel(7)^1xxcancel(6)^1xxcancel(5)^1xx4xx3xx2)cancel(10!)) #

#color(white)("ddddddddddd")(19xx13xx11)/(24) = 113.208#

#color(white)()#

#color(green)("Option 2")color(blue)(larr" A whole number !")#

#10! =10xx9xx8xx7xx6xx5xx4xx3xx2=3628800#

#color(white)()#

#color(green)("Option 3")color(red)(larr" Not a whole number !")#

#(10!)/2^10#

#(cancel(10)^5xx9xxcancel(8)^4xx7xxcancel(6)^3xx5xxcancel(4)^2xx3xxcancel(2))/ (cancel(2)xxcancel(2)xxcancel(2)xxcancel(2)xxcancel(2)xx2xx2xx2xx2xx2)#

#(5xx9xxcancel(2)^1xx7xx3xx5xx3)/(cancel(2)xx2xx2xx2xx2)=885.9375#

#color(white)()#

#color(green)("Option 4")color(red)(larr" Not a whole number !")#
#color(white)()#

#(9!)/2^10#

#(9xxcancel(8)xx7xxcancel(6)^3xx5xxcancel(4)xx3xxcancel(2))/(2xx2xx2xxcancel(2)xxcancel(2)xxcancel(2)xxcancel(2)xxcancel(2)xxcancel(2)xxcancel(2) )#

#(9xx7xx3xx5xx3)/(2^3) = 354.375#

#color(white)()#

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Let's have a look at the other answer provided by

Numeros fous #color(red)(larr" Not a whole number !")#

#(10(10+1)[2(10+1)])/6 = 10/3xx11^2 = 403 1/3#