Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

Find the number of arrangements of the letters of the word MADHUBANI. In how many of these arrangements
(i) Do the words start with B?
(ii) Do the vowels occur together?
(iii) Do your words begin with D and end in H?

seo-qna
Last updated date: 16th Apr 2024
Total views: 417.9k
Views today: 5.17k
MVSAT 2024
Answer
VerifiedVerified
417.9k+ views
Hint: - Here we go through by the rule of permutation and combination as we know the rule how the n letters are arranged at n places i.e. n! And when any number repeats by x times then we divide it by x!

Complete step-by-step answer:
Here in the question the given word is MADHUBANI.
There are 9 letters of which A appears twice and rest are all different in the word MADHUBANI and these 9 letters have to be arranged at 9 places so we arrange it by 9! But there A repeats two times so we divide it by 2!
$\therefore $ Required number of arrangements$ = \dfrac{{9!}}{{2!}} = \dfrac{{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1}} = 181440$ ways

(I) let us fix B at the extreme left positions then 8 letters are left, out of which A appears twice,by fixing B only 8 letters left and have to arrange at places so we arrange it by 8! But there A repeats two times so we divide it by 2!

$\therefore $Number of arrangements $ = \dfrac{{8!}}{{2!}} = \dfrac{{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1}} = 20160$ ways

 (ii) There are 3 vowels in the given word out of which A occurs twice. Since they have to always occur together, we treat them as a single object. This single object together with 6 remaining letters accounts for 7 letters. These can be arranged in 7! Ways. Corresponding to each of these arrangements the 3 vowels A, A and U are arranged in $\dfrac{{3!}}{{2!}}$
Hence, number of arrangements $ = 7! \times \dfrac{{3!}}{{2!}} = 15120$ ways

(iii) Let us fix D at left and H at the right end.
Remaining number of letters is 7 out of which A occurs twice.
$\therefore $ Required number of arrangements $ = \dfrac{{7!}}{{2!}} = \dfrac{{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1}} = 2520$ ways

Note:- Whenever we face such a type of question the key concept for solving the question is to go through the arranging method first do it as the question says then find out the letters which can be arranged at different places then apply the formula of arranging on that letters only to get our answers.