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Arrange the letters[every arrangement must contain all letters of the word] of the word 'BENGALI', so that no two vowels are together.
What my cute little brain could find out:
Let me first arrange the vowels...
__ E __ A __
I__
where the underscores contain the consonants. Now, clearly, there will be $^{3}P_3$ arrangements. So my brain tells me to find the ans for the E A I one and then multiply it by $^{3}P_3$
Now my brain thinks for a minute and then says:
"Hey! There are $4$ underscores and how many consonants do you have? Its $4$ Is it not a modified stars and bars problem?"
I thought for a moment, and agreed with my brain. Then it said:
"Find all integer solutions to the equation based on
the following conditions:
$x_1+x_2+x_3+x_4=4$, where $x_1,x_4≥0$ and $x_2,x_3≥1$"
And the answer to this is $\binom{2 + 4 - 1}{2} = \binom{2 + 4 - 1}{4 - 1}$[just some honesty!]
"But wait! There are $^{4}P_4$ ways of arranging the consonants. So multiply this by $^{4}P_4$"
And finally by $^{3}P_3$
So my
final answer is
$$\binom{2 + 4 - 1}{2}\times^{4}P_4 \times^{3}P_3$$ Am I correct? If yes, is there any better or more efficient way? If yes, would you show that?
asked Dec 10, 2017 at 17:18
ami_baami_ba
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There are $4$ consonants, hence $5$ slots to place one of the three vowels. The consonants as well as the vowels can be written in any order. It follows that there are $${5\choose3}\cdot 3!\cdot 4!=1440$$ admissible arrangements of the $7$ letters.
answered Dec 10, 2017 at 18:37
Christian BlatterChristian Blatter
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1
1] In what ways the letters of the word "RUMOUR" can be arranged?
- 180
- 150
- 200
- 230
Answer: D
Answer with the explanation:
The word RUMOUR consists of 6 words in which R and U are repeated twice.
Therefore, the required number of permutations =
Or,
Hence, 180 words can be formed by arranging the word RUMOUR.
2] In what ways the letters of the word "PUZZLE" can be arranged to form the different new words so that the vowels always come together?
- 280
- 450
- 630
- 120
Answer: D
Answer with the explanation:
The word PUZZLE has 6 different letters.
As per the question, the vowels should always come together.
Now, let the vowels UE as a single entity.
Therefore, the number of letters is 5 [PZZL = 4 + UE = 1]
Since the total number of letters = 4+1 = 5
So the arrangement would be in
5P5 =
Note: we know that 0! = 1
Now, the vowels UE can be arranged in 2 different ways, i.e., 2P2 = 2! = 2*1 = 2 ways
Hence, the new words, which can be formed after rearranging the letters = 120 *2 = 240
As we known z is occurring twice in the word ‘PUZZLE’ so we will divide the 240 by 2.
So, the no. of permutation will be = 240/2 = 120
3] In what ways can a group of 6 boys and 2 girls be made out of the total of 7 boys and 3 girls?
- 50
- 120
- 21
- 20
Answer: C
Answer with the explanation:
We know that nCr = nC[n-r]
The combination of 6 boys out of 7 and 2 girls out of 3 can be represented as 7C6 + 3C2
Therefore, the required number of ways = 7C6 * 3C2 = 7C[7-6] * 3C[3-2] =
Hence, in 21 ways the group of 6 boys and 2 girls can be made.
4] Out of a group of 7 boys and 6 girls, five boys are selected to form a team so that at least 3 boys are there on the team. In how many ways can it be done?
- 645
- 734
- 756
- 612
Answer: C
Answer with the explanation:
We may have 5 men only, 4 men and 1 woman, and 3 men and 2 women in the committee.
So, the combination will be
as we know that
nCr=
So, [7C3 * 6C2] + [7C4 * 6C1] + [7C5]
Or,
Or, 525 +210+21 = 756
So, there are 756 ways to form a committee.
5] A box contains 2 red balls, 3 black balls, and 4 white balls. Find the number of ways by which 3 balls can be drawn from the box in which at least 1 black ball should be present.
- 64
- 48
- 32
- 96
Answer: A
Answer with the explanation:
The possible combination could be [1 black ball and 2 non-black balls], [2 black balls and 1 non- black ball], and [only 3 black balls].
Therefore the
required number of combinations = [3C1 * 6C2] + [3C2 * 6C1] + [3C3]
r,
Permutation and Combination Concepts