- #1
juantheron
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How many permutation of the letters of the words $\bf{"MADHUBANI"}$ do not begin with $\bf{M}$ but end with $\bf{I}$
jacks said:How many permutation of the letters of the words $\bf{"MADHUBANI"}$ do not begin with $\bf{M}$ but end with $\bf{I}$
jacks said:How many permutation of the letters of the words $\bf{"MADHUBANI"}$ do not begin with $\bf{M}$ but end with $\bf{I}$
How many permutation of the letters of the words $\bf{"MADHUBANI"}$
do not begin with $\bf{M}$ but end with $\bf{I}$
Don't we need to divide [1] by 2! as well since we consider the cases when the A 's are switched to be the same?soroban said:
[1] The first letter is $A\!:\;A\,\_\,\_\,\_\,\_\,\_\,\_\,\_\,I $
. . .The remaining spaces can be filled in $7!$ ways.
Therefore, there are: .$7! + 6\left(\frac{7!}{2!}\right) \:=\:20,160$ ways.
Both methods are correct, but there is an error in soroban's computation.Jameson said:Don't we need to divide [1] by 2! as well since we consider the cases when the A 's are switched to be the same?
If so, then you'd have \(\displaystyle \frac{7!}{2!}+6\left(\frac{7!}{2!}\right)\) that is \(\displaystyle \left( \frac{7 \cdot 7!}{2!} \right)\) which what CB and I got.
Would like confirmation on this :)
That red 6 should be a 5: if the first letter is not I, M or either of the two As, then there are only five other choices (count them: D, H, U, B, N). So soroban's count for the number of arrangements should be $7!+5\bigl(\frac{7!}{2!}\bigr)$, which agrees with the answer of Jameson and CaptainBlack.soroban said:[2] The first letter is not $A$ and not $M\!:$
. . . $\_\,\_\,\_\,\_\,\_\,\_\,\_\,\_\,I$
There are 6 choices for the first letter.
. . . The remaining spaces can be filled in $\frac{7!}{2!}$ ways.
The permutations of "MADHUBANI" that do not start with "M" and end with "I" are as follows: "ADHUBANI", "ADHUBANI", "ADHUBANI", "ADHUBANI", "ADHUBANI", "ADHUBANI".
There are 6 possible permutations for "MADHUBANI" not starting with "M" and ending with "I".
One example of a permutation of "MADHUBANI" not starting with "M" and ending with "I" is "ADHUBANI".
Yes, the permutations follow a specific order based on the letters in the word. The first letter in the permutation will always be "A" and the last letter will always be "I".
The number of permutations can be calculated using the formula n!/(n-r)!, where n is the total number of letters in the word and r is the number of letters that are not changing positions (in this case, 2).