First we fix A as the first letter

No. of words possible with A as the first letter = 4! = 24

No. of words possible with M as the first letter= $\frac{4!}{2!}$ = 12

No. of words possible with RAA as the first 3 letters = 2

No. of words possible with RAM as the first 3 letters = 2

Next word will be RASAM

Rank= 41

$\begin{array}{cccccc}& X_1& X_2& X_3& 3X_4& =15\\ \text{Case 1:}& X_1& X_2& X_3& 3& =15\\ \text{Case 2:}& X_1& X_2& X_3& 6& =15\\ \text{Case 3:}& X_1& X_2& X_3& 9& =15\\ \text{Case 4:}& X_1& X_2& X_3& 12& =15\end{array}$
Case 1:
$X_1+X_2+X_3=12;$ Total number of natural number solution = ${}^{11}{C}_2=55$
Case 2:
$X_1+X_2+X_3=9;$ Total number of natural number solution = ${}^8{C}_2=28$
Case 3:
$X_1+X_2+X_3=6;$ Total number of natural number solution = ${}^5{C}_2=10$
Case 4:
$X_1+X_2+X_3=3;$ Total number of natural number solution = 1
Total number of natural number solution = 55 + 28 + 10 + 1 = 94

This is a question based on $S\stackrel{‘}{a}D$ with a varying lower limit.

Take out 1+2+3=6 from the RHS. Equation changes to

A+B+C = 4. Answer will the arrangement of 4 zeroes and 2 ones = ${}^6{C}_2$

No of ways in which the 3 correct envelopes can be selected= ${}^7{C}_3=35$

Derangement of 4 envelopes & letters = 9 (derangement value for 4 is 9)

Total No of ways of arrangement = $9\times 35$ = 315

To form a rectangle we need four intersecting lines.

The first two parallel lines can be selected in ${}^5{C}_2$ ways and the other two parallel lines can be selected in ${}^6{C}_2$ ways.

Total number of parallelograms which can be formed= $10\times 15=150$

If the first digit and last digit of the four digit number needs to be even, then we have to choose numbers out of 2, 6 and 8

The first digit can be chosen in 3 ways and the last digit in 2 ways = $3\times 2=6$ ways

The remaining two digits can be chosen and arranged among the remaining 4 digits in ${}^4{P}_2$ ways = 12 ways

Total number of ways = $6\times 12=72$ ways

0 balls are going to basket A. hence we will consider the distribution only in the remaining 3 baskets

B+C+D = 5

Total = ${}^7{C}_2=21$

If B, does not want any neighbour among D,E,F, B needs to be seated between only A and C. A and C can be arranged in 2! Ways

Similarly, E who does not want to be seated next to A,B or C will be seated between the other two( D,F). Those two can be arranged in 2! Ways.

Total number of arrangements= $2\times 2=4$

Use the concept of grouping, for a three digit number ABC

A+B+C = 10, A >1

It changes to A + B + C = 9

Number of possible solutions = ${}^{11}{C}_2=55$

Let’s number the baskets as A,B,C and D

With the first constraint, 2 of the balls are going to basket 2(B), hence we have 5-2=3 balls left to distribute among the remaining baskets.

A+C+D = 3

Number of distributions possible = ${}^5{C}_2$ = 10