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The teacher asked a class of fifth-grade students to find the average of 15 different two-digit numbers displayed on the blackboard. After the students finished the exercise, the teacher wrote the correct answer on the blackboard. Ankit, in an attempt to impress the teacher, claimed that he could reverse the digits of one of the numbers so that the new number also was a two-digit number, and the average he would get would be 2.4 more than the result displayed. What is the probability of Ankit finding such a number on the blackboard?

The teacher asked a class of fifth-grade students to find the average of 15 different two-digit numbers displayed on the blackboard. After the students finished the exercise, the teacher wrote the correct answer on the blackboard. Ankit, in an attempt to impress the teacher, claimed that he could reverse the digits of one of the numbers so that the new number also was a two-digit number, and the average he would get would be 2.4 more than the result displayed. What is the probability of Ankit finding such a number on the blackboard? Correct Answer 1 – (<span style="position: relative; line-height: 0; vertical-align: baseline; top: -0.5em;font-size:10.5px;">85</span>C<span style="position: relative; line-height: 0; vertical-align: baseline; bottom: -0.25em;font-size:10.5px;">15 </span>/ <span style="position: relative; line-height: 0; vertical-align: baseline; top: -0.5em; font-size:10.5px;">90</span>C<span style="position: relative; line-height: 0; vertical-align: baseline; bottom: -0.25em; font-size:10.5px;">15</span>) 

Calculation:

⇒ Since an average of the fifteen numbers has been increased by 2.4, therefore, the sum of the fifteen numbers Ankit has on his notebook must be greater than the actual sum by 15 × 2.4 = 36.

⇒ Since Ankit has left all the other fourteen numbers unaltered, that means that the new number formed by reversing the digits on the original number is greater than the original number by 36.

⇒ Let’s say that the original number was XY, therefore the new number will be YX.

⇒ Therefore, (10Y + X) – (10X + Y) = 36

Or, 9(Y – X) = 36

Or, Y – X = 4

⇒ Therefore, any number which does not have zero as one of its digits (the number cannot have zero since any number having zero when reversed will be turned to a one-digit number) and where the difference of the digits is 4 will help Ankit to achieve his feat.

⇒ 15, 26, 37, 48, 59 are the required number that can help Ankit.

⇒ Number of two digits numbers possible = 90

⇒ Number of ways in which 15 two digits number can be selected = 90C15

⇒ Number of ways in which 15 numbers can be selected such that the five numbers that can help Ankit in performing the mischief are not selected = 85C15

⇒ Hence, the probability that Ankit will find one of the given five numbers on the blackboard = 1 – (85C15 / 90C15). 


Additional Information

⇒The number of combinations of n things taken r (0 < r < n) at a time is given by nCr = n! / r!(n-r)!

⇒ The relationship between combinations and permutations is nCr = nPr / r!

⇒ The number of ways of selecting r objects from n different objects subject to a certain condition like:

1. k particular objects are always included =  n - kCr - k

2. k particular objects are never included =  n - kCr

⇒ The number of arrangement of n distinct objects taken r at a time so that k particular objects are

(i) Always included = n - kCr - k.r!,

(ii) Never included = n - kCr.r!.

⇒ In order to compute the combination of n distinct items taken r at a time wherein, the chances of occurrence of any item are not fixed and maybe one, twice, thrice, …. up to r times is given by n + r - 1Cr

⇒ If there are m men and n women (m > n) and they have to be seated or accommodated in a row in such a way that no two women sit together then total no. of such arrangements = m + 1Cn. m! This is also termed the Gap Method. 

⇒ If we have n different things taken r at a time in form of a garland or necklace, then the required number of arrangements is given by nCr(r - 1)!/2.

⇒ If there is a problem that requires n number of persons to be accommodated in such a way that a fixed number say ‘p’ are always together, then that particular set of p persons should be treated as one person. Hence, the total number of people in such a case becomes (n - m + 1). Therefore, the total number of possible arrangements is (n - m + 1)! m! This is also termed the String Method.  

⇒ Let there be n types of objects with each type containing at least r objects. Then the number of ways of arranging r objects in a row is nr. 

⇒ The number of selections from n different objects, taking at least one = nC1 + nC2 + nC3 + ... + nCn = 2n - 1.  

⇒ The total number of selections of zero or more objects from n identical objects is n +1.

Related Questions

The teacher asked a class of fifth grade students to find the average of 15 different two-digit numbers displayed on the blackboard. After the students finished the exercise, the teacher wrote the correct answer on the blackboard. Ankit, in an attempt to impress the teacher, claimed that he could reverse the digits of one of the numbers so that the new number also was a two-digit number, and the average he would get would be 2.4 more than the result displayed. What is the probability of Ankit finding such a number on the blackboard?
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With this exercise, I believe we can help students bridge their divides and replace anger and distrust with compassionate connections – just as I witnessed betweenShirley and Tiffany.
Shirley returned to class after a brief hiatus, keeping a cool distance from Tiffany. But over the weeks spent together they gradually came to know each other. They practiced seeing and listening, sharing stories so different that they felt bewildered as to how they could overcome the gap. But they found that acknowledging their differences led them to discover a place of deep connection in commonalities, such as being raised by grandmothers, and even wounds, including childhood trauma,that they never imagined existed.
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