Phrases Previous Year Questions (PYQs) – Permutation And Combination

We can get the triangles in two different ways.

Taking two points from the line having 10 points

(in ¹⁰C2 ways, i.e., 45 ways) and one point from the line consisting of 11 points (in 11 ways).

So, the number of triangles here is 45×11=495.

Taking two points from the line having 11 points (in ¹¹C2, i.e., 55 ways) and one point from the line consisting of 10 points (in 10 ways), the number of triangles here is 55×10=550

Total number of triangles, =495+550

=1,045 triangles

210=1*2*3*5*7=1*6*5*7. 

We have to make 5 digit no. And also divisible by 4 using digits (1,2,3,4,5,6) To be divisible by 4 last two digit of 5 digits no should be divisible by 4 so we choose last two digits st they are divisible by 4 and rest with remaining four digits. Last two digits only can be 12,16,24,32,36,52,56,64 as all are divisible by 4. So total no of ways= no ways of choosing first 3 digits* no ways of choosing last two digits Total= (4*3*2)*(8) =192

Possible cases:

• (5M,0F): ${}^7{C}_5{\times }^6{C}_0$
• (4M,1F): ${}^7{C}_4{\times }^6{C}_1$
• (3M,2F): ${}^7{C}_3{\times }^6{C}_2$

Total = ${}^7{C}_5{\times }^6{C}_0{+}^7{C}_4{\times }^6{C}_1{+}^7{C}_3{\times }^6{C}_2$ $=21+210+525=756$

Answer: $\boxed{{\displaystyle 756}}$ ✅