Question

The number of words that can be formed using any number of letters of the word "KANPUR" without repeating any letter is

• A. $720$
• B. $1956$
• C. $360$
• D. $370$

Answer 1

Answer: (B)

$1956$

$\left(a\right)1$ Letter word:

Selecting $1$ letter of $6$ letter $K,A,N,P,U,R={}^6{C}_1=6$ ways

$\left(b\right)2$ Letter words:

Selecting $2$ letter of $6$ letter $K,A,N,P,U,R={}^6{C}_2$

Therefore no. of ways to create $2$ letter words $={}^6{C}_2\times 2!$

$\left(c\right)3$ Letter words:

Selecting $3$ letter of $6$ letter $K,A,N,P,U,R={}^6{C}_3$

Therefore no. of ways to create $3$ letter words $={}^6{C}_3\times 3!$

$\left(d\right)4$ Letter words:

Selecting $4$ letter of $6$ letter $K,A,N,P,U,R={}^6{C}_4$

Therefore no. of ways to create $4$ letter words $={}^6{C}_4\times 4!$

$\left(e\right)5$ Letter words:

Selecting $5$ letter of $6$ letter $K,A,N,P,U,R={}^6{C}_5$

Therefore no. of ways to create $5$ letter words $={}^6{C}_5\times 5!$

$\left(f\right)6$ Letter words:

Selecting $6$ letter of $6$ letter $K,A,N,P,U,R={}^6{C}_6$

Therefore no. of ways to create $6$ letter words $={}^6{C}_6\times 6!$

Total $=6+{}^6{C}_2\times 2!+{}^6{C}_3\times 3!+{}^6{C}_4\times 4!+{}^6{C}_5\times 5!+6!=1956$ words