Question of the Week - 40 (The set S is defined as {1, -2, -3, 8,...)

What is this CHI anyways ?

Could not understand this ...

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This is just a term used to frame the question. You can also consider this as a function which gives the final output as the product of any 3 elements present in the given set S.

Product of three elements in the set S is defined as chi

We have to find the cases where chi>=7

Total negative elements = 5
Total positive elements = 5

Following are the only possibilities when product of three elements >= 7

1) 2 negative and 1 positive - 2 negative out of 5 can be selected in 5C2 ways and 1 positive in 5 ways

Number of possible values of chi = 5C2 * 5 - 1 [subtracting for the case when -2,-3 and 1 will be selected]
=50 -1 = 49
2) three positive terms

Out of 5 three positive can be chosen in 5C3 ways
Number of possible values = 5C3 = 10

For every other combination such as , (3 Negative , 2 Positive + 1 negative)value of chi will be negative.

Hence possible values of chi = 49 + 10 =59

IMO (B)

yes , but when choosing (-2 , -3 , positive number other 1 ) isn't this the same when we choose (6 , 1 , other positive number ) let's say we choose the following
-2 , -3 , 30
6 , 1 , 30
both have the same result and we counted them twice , same for -2 , -3 , 8 and 6 , 1 , 8
also : -22 , 3 , negative other than -22 or -11
and 6 , -11 , negative other than -22 or -11

I'm not gonna count all possible double entrees since the only answer below 59 is is A
please correct me

I think you are right.But I could count only 3 values for which it holds true (-2,-3,8);(-2,-3,21);(-2,-3,30) with (6,1,8);(6,1,21);(6,1,30)

So number of values: (59-3) = 56



3 is not in the set.

other combinations will not be possible :- (-22,-3) with (1,8,21,30,6) and (6,-11) with (-2,-3,-22,-5) no values overlap.

I could not find other overlaps. Could not count it down to 50. Can you please list if you find other 6 numbers?

Solution

Given:

• Elements of set S are {1, -2, -3, 8, 21, 30, -22, -5, 6, -11}.
• chi of set is defined as the product of any 3 elements of the given set S.

To find:

• The number of values of Chi >=7 for set S.

Approach and Working:

• The number of values of Chi being at least 7 = The number of values of Chi being greater 0 - The number of values of Chi from 1 to 6.

Number of ways we can get chi >0

• Chi will be greater than 0 when product of three numbers is positive.

o That can happen in two cases when we multiply 3 numbers.

1. All number positive.

• Ways to select three positive numbers= 5c3 =10
2. Two negative and one positive.

• Ways to select two negative numbers * Ways to select one positive number = 5c2 * 5c1 = 10 *5 =50
o Number of ways in which chi >0= 10 +50 =60

Number of ways we can get chi from 1 to 6.

• We cannot directly know when chi can be from 1 to 6.
• So, we will first calculate the least positive values of Chi and then second least and so on.
o
Least value of chi can be either the multiplication of least three positive integers or the multiplication of two greatest negative integers and one least positive integer.

 Multiplication of three minimum positive integers = 1 * 6 *8
• This is certainly greater than 7.

 Multiplication of two greatest negative integers and one least positive integer = -2 * -3 *1 =6
• So, least value of chi =6
o Second least value of chi is greater than 7.
• Hence, Number of values of chi from 1 to 6 =1

The number of values of chi being at least 6 = 60-1=59.

Correct Answer: B

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