1. Calculate the number of integers divisible by 4 between 50 and 500, inclusive

Question

1. Calculate the number of integers divisible by 4 between 50 and 500, inclusive.

2. Hexadecimal digits are formed using either a numeric decimal digit or a letter from A to F. How many possible digits can be chosen? Did you use the sum rule or the product rule?

3. A combination lock requires three numbers from 1 to 25. How many combinations are possible? Did you use the sum rule or the product rule?

4. Use the permutation formula to calculate the number permutations of the set {a, b, c, d} taken two at a time. Also list these permutations.

5. Using the pigeonhole principle, show that any 11 digit decimal number must contain two instances of some decimal digit.

Explanation / Answer

Ans:

[1] First number divisible by 4 between 50 and 500 is 52

Second number divisible by 4 between 50 and 500 is 56

Third number divisible by 4 between 50 and 500 is 60

. ........ and so on.....Including last number 500 which is also divisible by 4.

This forms an arithematic sequence i,e,

52+56+60+64+..........+500

Using the formula for the sum of arithmatic sequence:

nth term = FirstTerm + CommonDifference * (NumberOfTerms - 1)

Therefore,

nth term = 500, FirstTerm = 52, CommonDifference = 56 - 52 = 4, NumberOfTerms = n

500 = 52 + 4*(n-1)

500 = 52 + 4n - 4

n = 113 Ans.

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[4]

Given,

set = {a, b, c, d}

Number of elements in the set = 4

Number of favourable events needed to be occur according to question = 2

Therefore,

Number of permutation will be:

[4 P 2] = [4 !] / [2 !] = 4 * 3 = 12

List of permutations: {(a,b), (a,c), (a,d), (b,a), (b,c), (b,d), (c,a), (c,b), (c,d), (d,a), (d,b), (d,c)} Ans.

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[2]

According to the sum rule, if Task A can be done in m ways & Task B can be done in n ways.Then either of the tasks can be done in (m+n) ways.

Task A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 0}

Task B = {A, B, C, D, E, F}

Therefore, total number of possible digits that can be choosen: (m+n) = (10+6) = 16 Ans.

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[5]

We have 10 digits from 0 to 9, Therefore in order to make any 11 digit decimal number there must contain atleast two instances of any digit from 0 to 9.

Pigeon Hole principle clearly states that if there are n boxes and n+1 pigeon, so there must be a box that contain more that one pigeon in order to fit the pigeos into the boxes.

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