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Solution

A number divisible by 5 must end in 5.

Fix the units digit as 5.

Remaining digits for hundreds and tens places: 2, 3, 6, 7, 9 (5 digits).

Number of ways to choose hundreds and tens digits = 5 × 4 = 20.

Answer: 20

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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

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210=1*2*3*5*7=1*6*5*7. 

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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

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Solution

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Possible cases:

• (5M,0F): $^7C_5 \times ^6C_0$
• (4M,1F): $^7C_4 \times ^6C_1$
• (3M,2F): $^7C_3 \times ^6C_2$

Total = $^7C_5\times^6C_0 + ^7C_4\times^6C_1 + ^7C_3\times^6C_2 $ $= 21 + 210 + 525 = 756$

Answer: $\boxed{756}$ ✅

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Solution

Stations = 7 (Tatanagar + 5 stops + Howrah) Number of one-way tickets = number of ordered pairs = n(n−1)/2 = 7×6/2 = 21 Correct answer: 21

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Solution

Task 2 has 2 choices (person 3 or 4). Casework gives total 180 valid permutations.