Aptitude - Permutation and Combination - Discussion

Yeah, it is correct but there is a simpler way to do this;

Out of 11 places, the last 4 take it as vowels so there would be 4! places and the remaining 7 places will come together as 7! ways.

Hence, 7!*4!=120960 (that's the answer).

GIVEN WORD = M A T H E M A T I C S = TOTAL = 11.

Solution:-

VOWEL FROM THE WORD = A A E I -------------------------> (1)
REMAINING LETTERS IN THE WORD = M T H M T C S ------->(2)

CONSIDER [ A A E I ] M T H M T C S = 1 + 7 = 8! divided by 2! 2! [becus we have M M and T T].

FROM ------->A A E I we get 4! and divided 2! [becus we have two A A].

8! 4! DIVIDED BY 2! 2! 2! =======> 120960.

Anyone can please solve this for me? In how many ways PENCIL be arranged so that P and C are next to each other?

In the word 'MATHEMATICS', we treat the vowels AEAI as one letter.

Thus, we have MTHMTCS (AEAI).

7! * 4! => 5040 * 24 => 120960.

Make it simple!

MATHEMATICS.

# The consonants MTHMTCS are considered as 7 letters.

# The four vowels AEAI are considered as one letter since they are supposed to be arranged all together.

# The 7 consonant + 1 group of vowels= 8 letters.

# If there are no repeated letters in these 8 letters, the total ways of arrangements would be 8 != 40320.

# But in the 8 letters we set up above we know M and T are happened to occur twice. Since a letter repeated twice has two arrangements M and T add up to have 4 repeated arrangements. These 4 repeated arrangements will divide the total 40320 arrangements in to 4. Hence, 40320÷4= 10080.

# The AEAI letter will form (4!) 24 arrangements if there is no letter repeated. But A is repeated twice which divide the 24 arrangements by 2. Hence, we do have 12. arrangements.

# Finally the total number of ways to arrange is፡.

10080*12=120960 ways.

My answer is 11!/2!2!2!. Is it is wrong, please explain the answer I have a dought.

I am not getting this answer please sir give me a clear explanation.

@ALL.

We should notice that in the word "mathematics'", m and t occurred twice 8!/2!2!.

In the word MATHEMATICS, We treat the vowel as one letter. Thus we have MTHMTCS (AEAI). Now we have t arrange 7 letters, out of which M and T occur twice. So no of arranging these letters = 7!/2!*2! = 1260.

Now, AEAI has 4 letters in which A occurs twice.

So, no of ways of arranging these letters = 4!/2!=12.
Required no of words= 1260*12=15120.
So answer is none of these.

If vowels never come together then?