#2 Permutations and Combinations

There are 5!/2!=60 ways to arrange all the letters. Among them, 12 ways start with N. Thus, 50th­ is the 2nd with N at front, that is NAAIG

There are 5 letters in the word AGAIN , in which A appears 2 times. Therefore, number of words can be formed from letters of AGAIN= ( $5\times 4\times 3 \times 2\times 1$ )/( $2\times 1$ ) .

To get the number of words starting with A , we fix the letter A at the extreme left position, we then rearrange the remaining 4 letters taken all at a time. There will be as many arrangements of these 4 letters taken 4 at a time as there are permutations of 4 different things taken 4 at a time. Hence, the number of words starting with A = $4\times 3 \times 2\times 1$ )/( $2\times 1$ )= 24 .

Then tarting with G ,The number of words=( $4\times 3 \times 2\times 1$ )/( $2\times 1$ )= 12 as after replacing G at the extreme left position, we are left with letters A , A , I and N . Similarly, there are 12 words starting with the next letter I . Total number of words so far obtained = 24+12+12 = 48 .

The $49^{th}$ word is NAAGI . The $50^{th}$ word is NAAIG .