In the word 'MATHEMATICS', we treat the vowels AEAI as one letter.
Thus, we have MTHMTCS (AEAI).
Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different.
Number of ways of arranging these letters = 8!/(2! x 2!)= 10080.
Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.
Number of ways of arranging these letters =4!/2!= 12.
Required number of words = (10080 x 12) = 120960
Required number of ways= = ( )= 11760.
The bus fromA to B can be selected in 3 ways.
The bus from B to C can be selected in 4 ways.
The bus from C toD can be selected in 2 ways.
The bus fromD to E can be selected in 3 ways.
So, by the General Counting Principle, one can travel fromA to E in 3 x 4 x 2 x 3 ways = 72
A person can be chosen out of 18 people in 18 ways to be seated between Musharraf and Manmohan. Now consider Musharraf, Manmohan, and the third person, sitting between them, as a single personality, we can arrange them in 17! ways but Musharraf and Manmohan can also be arranged in 2 ways.
Required number of permutations = 18 x (17!) x 2 = 2 x 18!
n(S) = 52C3 = 132600/6 = 22100
n(E) = 4C3 = 24/6 = 4
Here given the required digit number is 4 digit.
It must be divisible by 5. Hence, the unit's digit in the required 4 digit number must be 0 or 5. But here only 5 is available.
x x x 5
The remaining places can be filled by remaining digits as 5 x 4 x 3 ways.
Hence, number 4-digit numbers can be formed are 5 x 4 x 3 = 20 x 3 = 60.
Out of 26 alphabets two distinct letters can be chosen in ways. Coming to numbers part, there are 10 ways.(any number from 0 to 9 can be chosen) to choose the first digit and similarly another 10ways to choose the second digit. Hence there are totally 10X10 = 100 ways.
Combined with letters there are X 100 ways = 65000 ways to choose vehicle numbers.
For any set of 4 points we get a cyclic quadrilateral. Number of ways of choosing 4 points out of 12 points is = 495.
Therefore, we can draw 495 quadrilaterals
Since order does not matter, use the combination formula
= 24/6 = 4
If a card hand that consists of four Queens and an Ace is rearranged, nothing has changed.
The hand still contains four Queens and an Ace. Thus, use the combination formula for problems with cards.
We have 4 eights and 4 sevens.
We want 3 eights and 2 sevens.
C(have 4 eights, want 3 eights) x C(have 4 sevens, want 2 sevens)
C(4,3) x C(4,2) = 24
Therefore there are 24 different ways in which to deal the desired hand.
ABACUS is a 6 letter word with 3 of the letters being vowels.
If the 3 vowels have to appear together, then there will 3 other consonants and a set of 3 vowels together.
These 4 elements can be rearranged in 4! Ways.
The 3 vowels can rearrange amongst themselves in 3!/2! ways as "a" appears twice.
Hence, the total number of rearrangements in which the vowels appear together are (4! x 3!)/2!
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