InfyTQ Permutation and Combination Quiz 4

No of letters in "PLAY" is 4

No of Arrangements = 4! = 24

He can pay by all 10 rupee notes in
5 days = 1 way
3 Ten rupee + 1 twenty rupee
= \frac{4!}{3!\times 1!} = 4 ways
1 Ten rupee + 2 twenty rupee notes
= \frac{3!}{2!\times 1!} = 3 ways
Total ways = 1 + 4 + 3 = 8

If the we have to form even numbers,
units digit must be 2, 4, 6. i.e., 3 ways.
Also 5th digit should be even.
So it can be filled in 2 ways.
Now remaining 5 digits can be
filled in 5! ways.
So total 5! × 3 × 2 = 720 ways.

3600 = 2^{4} \times 3^{2} \times 5 ^{2}
Let abc = 2^{4} \times 3^{2} \times 5 ^{2}
We have to distribute four 2's to
three numbers a, b, c in

^{4+3-1}\textrm{C}_{3-1}

=^{6}\textrm{C}_{2}

= 15 ways.
Now two 3's has to be distributed
to three numbers in ^{2+3-1}\textrm{C}_{3-1}=

^{4}\textrm{C}_{2}

= 6 ways
Now two 5's has to be distributed
to three numbers in

^{2+3-1}\textrm{C}_{3-1}

=^{4}\textrm{C}_{2}

= 6 ways
Total ways = 15 × 6 × 6 = 540

If left and right digits are fixed with 5 and 6,
then the remaining 5 places can be filled by remaining 8 digits in

^{8}\textrm{P}_{5}

= 6720 ways.

The given data indicates that 14 points are collinear and remaining 22 points are non collinear.
A triangle can be formed by taking 1 points from 14 and 2 points from 22 (or)
2 points from 14 and 1 points from
22 (or) 3 points from 22
⇒ ^{14}\textrm{C}_{1}\times ^{22}\textrm{C}_{2}+^{14}\textrm{C}_{2}\times ^{22}\textrm{C}_{1}+^{22}\textrm{C}_{3} = 6776

 

Following would possible scenarios: Men - (0, 1, 2, 3)

If there are 0 men, then 11 women. Similarly, if there are 1 men then there would be 10 women so on ...

• Ways of choosing (0 Men, 11 Women) : (⁵C₀ × ¹¹C₁₁)


• Ways of choosing (1 Men, 10 Women) : (⁵C₁ × ¹¹C₁₀)


• Ways of choosing (2 Men, 9 Women) : (⁵C₂ × ¹¹C₉)


• Ways of choosing (3 Men, 8 Women) : (⁵C₃ × ¹¹C₈)

Atleast six means it can be 6 or more than so
No of ways :
First way 6W,4M is ^{10}\textrm{C}_{6}\times ^{10}\textrm{C}_{4} =44100
Second way 7W,3M is ^{10}\textrm{C}_{7}\times ^{10}\textrm{C}_{3} =14400
Third way 8W,2M is ^{10}\textrm{C}_{8}\times ^{10}\textrm{C}_{2} =2025
Fourth way 9W1M is ^{10}\textrm{C}_{9}\times ^{10}\textrm{C}_{1} =100
Fifth way 10W 0M is ^{10}\textrm{C}_{10}\times ^{10}\textrm{C}_{0} =1
Total no of ways is adding the above possible
ways we get 60626

let number of direct roads between A and B = x
B and C = y
C and A = z
x+zy = 33 .......(1)
y+xz = 23 .........(2)
on solving above equations we get z = \frac{10}{(y-x)} +1
z must be an integer as it indicates number of roads
so y-x can be 1,2,5,10
for y-x =2 equations (1) and (2) are satisfied
=> z = 6

Number of digits = 4

Total possible number of arrangement = 4 x 4 x 4 x 4 = 256

 

Group { 1, 2, 3, 4 }  as 1 set

⇒ Number of sets = 256 ÷ 4 = 64

 

Value of 1 set = 1 + 2 + 3 + 4 = 10

⇒ Value of 64 sets = 10 x 64 = 640

 

At ones place, the sum of all the digits = 640 x 1 = 640

At tens place, the sum of all the digits = 640 x 10 = 6400

At hundreds place, the sum of all the digits = 640 x 100 = 64000

At thousands place, the sum of all the digits = 640 x 1000 = 640000

 

Total sum = 640000 + 64000 + 6400 + 640 = 711040