Option (c)

If someone is going from P to Q. He has to travel 4 times vertically and 5 times horizontally. Ultimately, we are arranging 4 - V's and 5 - H's .

Number of ways to arrange 4 - V's and 5 - H's = 9!/(4! x 5!) = ⁹C_{4 = ⁹C5}

Or, There are 4 rows and 5 columns. Using the one-zero method, number of paths will be ⁹C₄ = 126 ways

Suppose there were no restrictions whatsoever and the problem simply said "How can you distribute k prizes among p students?".

 For each prize, there are p possibilities for the recipient of that prize. In total, that means there are p⋅p⋅.....⋅p (with k copies of p in the product) possible ways to assign the prizes. This is the p^k part.

or, we can say.  This is a type of “Different to Different” question.

5 prizes can be distributed among 4 students in 4⁵ ways.

Answer: Option C. ->

We need to arrange 2 Cs, O, 3Ns, V, 3Es and I. This is permutation with repetition.

It can be done in = 20160

Hence option (c)

Seventh term of the AP = a + 6d

Eleventh term of the AP = a + 10d

7(a + 6d) = 11(a + 10d)

-> 7a + 42d = 11a + 110d

-> 4a + 68d = 0

-> a + 17 d = 0

So, the eighteenth term is zero. Hence option (b)

As, we know sum of the n terms in an A.P.

(S_n)

In the given question S_n = 720 ;a = 12; d = 17; n =?

720 =

Solving we get n = 9. 9 terms are needed to get the same

Hence option (d)

Answer: Option E. ->  

NN_ _ _ _ _ _ _EE

The 7 places have to be filled by 2 Cs, O, N, V, E and I. This can be done in 7!/(2!)

Hence option (e)

p_th term = a + (p – 1) d = q ... (i)

q _th term = a + (q – 1) d = p ... (ii)

Subtracting (ii) from (i)

(p – 1) d – (q – 1) d = q – p

->pd – d – qd + d = q – p

->d (p – q) = q – p

-> d = -1a = q – (p – 1)d = q + (p – 1)

So, m_th term = q + p – 1 – (m – 1) = q + p – 1-m + 1 = p + q – m

Hence option (b)

Alternatively, use assumption

Let the first term=2 and the 2^nd term=1, therefore p=1 and q=2

Let m=3

Thus, 3^rd term=0. Substitute the values of p,q and m in the answer options and look for 0. Only option (b) gives this value

When none of the digits are repeated:

The hundred’s place can be filled by any of the digits: 2, 3, 5, 6, 7 or 8 except the one which has already been used at the thousand’s place, so it can be filled in 5 ways.

Similarly tens’ place can be filled in 4 ways: only those 4 numbers which have not been use either at hundred’s or thousand’s place.

Unit’s place can be filled in only 3 ways.

So, total number of nos. Possible = 4* 5*4*3 = 240

Hence option (b)

When the repetition is allowed:

All the places: Hundred’s, ten’s and unit’s can be filled by any of the digits: 2, 3, 5, 6, 7 or 8 i.e. in 6 ways. Thousand’s place can be filled only in 4 ways. Hence no. numbers possible = 4*6*6*6 = 864.

Hence option (d)