Math, asked by parvindermast, 4 months ago

Let m is number of words that can
be formed using all 10 letters of
word 'SIDHESHWAR such that all
words begin with 'SSHH' and n is
number of 6 lettered words that
can be formed using all letters of
m
word "VRINDA", then is
n
plz answer correctly otherwise I will report​

Answers

Answered by SrijanShrivastava
3

 \boxed{ 1}\boxed{ 1}\boxed{ 1}\boxed{ 1}\boxed{ 6}\boxed{ 5}\boxed{ 4}\boxed{ 3}\boxed{ 2}\boxed{ 1}\\ \boxed{ \small{S}} \boxed{\small{S}}\boxed{\small{H} }\boxed{\small{H}}\boxed{\small{?}}\boxed{ \small{?}}\boxed{ \small{?}}\boxed{ \small{?}}\boxed{\small{ ?}}\boxed{ \small{?}}

Thus the Number of ways of Arranging The rest 6 letters are 6! = 720 ways

\boxed{ 6}\boxed{ 5}\boxed{ 4}\boxed{ 3}\boxed{ 2}\boxed{ 1}

Similarly, Number of ways = 6! = 720

Answered by Rameshjangid
0

Answer:

To form a 6 lettered word using all letters of 'SIDHESHWAR', we can select 6 letters out of 10 and arrange them. The number of ways to select 6 letters out of 10 is 10C6 = 210, and the number of ways to arrange them is 6!/(2!1!1!1!1!) = 360.

Therefore, the value of n is 360.

Explanation:

The given word is 'SIDHESHWAR', and we need to form words that begin with 'SSHH' using all 10 letters of the word.

First, we need to find the number of ways to arrange the remaining 6 letters after 'SSHH'. There are 6 letters remaining in the word 'SIDHESHWAR', and the number of ways to arrange them is 6!/(2!2!1!1!)= 180. Here, we divided by 2! twice because there are two S's and two H's in the remaining 6 letters.

Next, we need to add the letters 'SSHH' at the beginning of each of these arrangements. So, the total number of words that can be formed is 180.

Now, we need to find the number of 6 lettered words that can be formed using all letters of the word 'SIDHESHWAR'. The word 'VRINDA' is not used in this question.

To form a 6 lettered word using all letters of 'SIDHESHWAR', we can select 6 letters out of 10 and arrange them. The number of ways to select 6 letters out of 10 is 10C6 = 210, and the number of ways to arrange them is 6!/(2!1!1!1!1!) = 360.

Therefore, the value of n is 360.

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