Quiz 2

Let there be g girls and b boys.

Number of games between two girls = gC2 Number of games between two boys = bC2 ∴ g(g – 1)/2 = 45 ∴ g2 – g – 90 = 0 ∴ (g – 10)(g + 9) = 0 ∴ g = 10

Also, b(b – 1)/2 = 190 ∴ b2 – b – 380 = 0 ∴ (b + 19)(b – 20) = 0 ∴ b = 20 ∴ Total number of games = (g + b)C2 = 30C2 = 435 ∴ Number of games in which one player is a boy and the other is a girl = 435 – 45 – 190 = 200 Hence, option 1.

Solution:

The number of ways of selecting 1 black square is 32.  

8 white squares in the corresponding row and column cannot be selected

The number of ways of selecting the white square is 24.  

Total number of required ways = 32 × 24 = 768  

Hence, option 4

The number of ways in which this can be done is 11*10*9*8 = 7920

26*25*24-15*14*13=12870.....here total 26 words and the password has three words...so total passwords are 26*25*24.but we have to minus asymmetric words password.so minus 15*14*13...answer is 12870.

Case 1: only 1 box with green ball. Number of possible ways = 6

 

Case 2: 2 boxes with green balls

Table shows that the green balls are placed in the boxes which are consecutively numbered. Number of possible ways = 5

 

Similarly,

Case 3: 3 boxes with green balls

Number of possible ways = 4

 

Case 4: 4 boxes with green balls

Number of possible ways = 3

 

Case 5: 5 boxes with green balls

Number of possible ways = 2

 

Case 6: 6 boxes with green balls

Number of possible ways = 1

? Total number of ways = 6 + 5 + 4 + 3 + 2 + 1 = 21 Hence, option 2.

If we have m lines in the north direction and n lines in west direction then the
total number of ways to move from one end to the diagonally opposite end
is given by (m + n – 2) Cn–1
∴ Required answer is (6 + 4 – 2)C4–1 = 8
C3
= 56. Ans.(2)

Let there be g girls and b boys.
Number of games between two girls = gC2
Number of games between two boys = bC2
∴ g(g – 1)/2 = 45
∴ g2 – g – 90 = 0
∴ (g – 10)(g + 9) = 0
∴ g = 10
Also,
b(b – 1)/2 = 190
∴ b2 – b – 380 = 0
∴ (b + 19)(b – 20) = 0
∴ b = 20
∴ Total number of games = (g + b)C2 = 30C2 = 435
∴ Number of games in which one player is a boy and the other is a girl = 435 – 45
– 190 = 200
Hence, option 1.

ANS 8

(100c2 X2!)/2!=4950

Ans: $6$

Assume Vaibhav wrote $n$ positive prime numbers.

Then, there are $\text{nC3}$ possible triplets. Vikram wrote product of all these triplets. (i.e., there are $\text{nC3}$ products he wrote.)

For every pair of these products, Vishal wrote corresponding GCD in which there are $90$prime numbers. Let\'s figure out a way to represent this $90$ in terms of $n$ so as to find out $n$

There are $4$ cases.  (for the sake of explanation, suppose the numbers written by Vaibhav are $a,b,c,d,e,f,\cdots$)

Case 1:  GCD of products like $(a\times b\times c)$ and $(d\times e\times f)$ (no number in common)
In this case, GCD$=1$ and not a prime.

Case 2:  GCD of products like $(a\times b\times c)$ and $(a\times d\times e)$ (one number in common)
In this case, GCD$=a$ and a prime.

Case 3:  GCD of products like $(a\times b\times c)$ and $(a\times b\times d)$ (two number in common)
In this case, GCD$=a\times b$ and not a prime.

Case 4:  GCD of products like $(a\times b\times c)$ and $(a\times b\times c)$ (three number in common)
In this case, GCD$=a\times b\times c$ and not a prime.

Therefore, our count $90$ only include GCD of products mentioned under case 2.

In other words, $(a\times b\times c)$ and $(a\times d\times e)$ represent one pair and there are $90$ such pairs where exactly one number is common.

Since there are $n$ numbers, the first number (like $a$) can be selected in $\text{nC1}=n$ ways.
Next two numbers (like $b,c$) can be selected in $(n-1)\text{C}2$ ways.
Next two numbers (like $d,e$) can be selected in $(n-3)\text{C}2$ ways.

Therefore, total such pairs $=\frac{n\times (n-1)\text{C}2\times (n-3)\text{C}2}{2!}$
(We divided by $2!$ to avoid overcounting.)

Now we have,
$\frac{n\times (n-1)\text{C}2\times (n-3)\text{C}2}{2!}=90\Rightarrow n\times (n-1)\text{C}2\times (n-3)\text{C}2=180\Rightarrow n\times \frac{(n-1)(n-2)}{2!}\times \frac{(n-3)(n-4)}{2!}=180\Rightarrow n(n-1)(n-2)(n-3)(n-4)=720\Rightarrow n=6$