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Q #16
:

In how many ways, we can choose a black and a white square on a chess board such that the two are not in the same row or column?

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Number of ways a black square can be chosen = 32.

Number of ways a white square can be chosen which is not in same

row or column as the black square = 24.

So the total number of ways = 32 × 24 = 768.


Q #17
:

Let n be the number of different 5 digit numbers, divisible by 4 with the digits 1, 2, 3, 4, 5 and 6, no digit being repeated in the numbers. What is the value of n?

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Last two digit should be 12, 16, 24, 32, 36, 52, 56, 64 – 8 ways

Remaining three places 4 × 3 × 2 = 24

Total 8 × 24 = 192 ways.


Q #18
:

Sam has forgotten his friend’s seven-digit telephone number. He remembers the following : thefirst three digits are either 635 or 674, the number is odd, and the number nine appears once. If Sam were to use a trial and error process to reach his friend, what is the maximum number of trials he has to make to succeed?

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he phone number is of the form

635 _ _ _ _ or 674 _ _ _ _

and the digit 9 is used exactly once.

So the problem is reduced to identifying a 4 digit odd number which uses 9 exactly once.

_ _ _ _

when 9 is at the thousand's or hundreds or ten's place, the other three digits can be filled in 9, 9 and 4 (for

units place) ways respectively.

 

No. of numbers

and when 9 is at unit's place, the number is definitely odd. The total numbers possible is this case is

total number of number that can be formed

= 4 × 9 × 9 × 3 + 9 × 9 × 9 = 1701

but 1701 numbers are possible for both telephone numbers starting with 635 and 674.

Total number of tries

= 2 × 1701 = 3402 .


Q #19
:

A palindrome is a number that reads the same left to right as it does from right to left, such as 252. How many six-digit palindromes are there which are even?

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Since we need 6 digit palindrome we only need to identify 3 digits

– – – / – – –

out of these three, digit at unit’s place can be selected in 4 ways (since it should be an even number and 0 cannot come) and other two digits can be selected in 10 ways so total number of ways

10 × 10 × 4 = 400.


Q #20
:

What will be the sum of all the four-digit numbers that can be made with the digits 0, 1, 2 and 4?

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Consider when 4 comes at unit's place.

total number of 4 digit numbers that can be formed

4

2 2 1

= 2 × 2 = 4

Thus, the digit 4 will come at the unit's place 4 times.

Same can be said of digits 1 and 2.

sum of unit's digit column (when all possible numbers of 4 digit using 0, 1, 2 are written one below the

other)

= 4 × 4 + 4 × 2 + 4 × 1 = 28

same will be for hundred's and ten's column.

In case of 1, 2 or 4 coming in thousand's place, total number of numbers that can be formed

4

3 2 1

= 3 × 2 = 6

in thousand's column, 4 will come 6 times and so will and 2.

We can now write the sum as.

Thousand’s

4 5

6×4

6×2

6×1

Hundred’s

3 1

4×4

4×2

4×1

Ten’s

3 0

4×4

4×2

4×1

One’s

2 8

4×4

4×2

4×1

3 3 2 (note: zeros may

come in any column

any number of

times and still will

not affect the sum)

Thus the answer is 45108.