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Question: Find the number of ways in which 6 red roses and 3 white roses of different sizes can be made to for...

Find the number of ways in which 6 red roses and 3 white roses of different sizes can be made to form a garland so that
(a) no two white roses came together

Explanation

Solution

Here we use the concept that garland is a circular decorative material made with flowers. Given two kinds of flowers we find ways to place each kind of flowers in the garland.
We first find a number of ways to place red flowers in a garland using the formula of circular permutation. Then taking the number of vacant spaces between red flowers as n and number of white flowers as r we find the number of ways to place white flowers using a combination formula. Multiply the number of ways in which 3 white flowers can be arranged within themselves using permutation method.

  • Number of ways to arrange n items in a circular manner is given by (n−1)!(n - 1)!
  • Number of ways to choose r items out of n items is given by combination formula which is given by nCr=n!(n−r)!r!^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}

Complete step-by-step answer:
We draw a rough diagram of the situation where we first place red flowers in a circular way and then find a number of places where we can place white flowers. Let the red flowers be denoted by red color empty spaces between the red flowers be denoted by white circles.

Number of red roses is 6
Number of white roses is 3
We use formula for circular permutation i.e. (n−1)!(n - 1)!
Number of ways to place 6 red flowers in circular way is =(6−1)! = (6 - 1)!
Therefore, number of ways to place 6 red flowers is 5!5!
Therefore, number of ways to place 6 red flowers is 5×4×3×2×15 \times 4 \times 3 \times 2 \times 1
Therefore, number of ways to place 6 red flowers is 120120 … (1)
Now we have 6 empty spaces between 6 flowers.
We have to place 3 white flowers in 6 empty places.
We use combination formulanCr=n!(n−r)!r!^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}
⇒\RightarrowNumber of ways to place 3 white flowers in 6 positions =6C3{ = ^6}{C_3}
Use the expansionnCr=n!(n−r)!r!^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}
⇒\RightarrowNumber of ways to place 3 white flowers in 6 positions=6!(6−3)!3! = \dfrac{{6!}}{{(6 - 3)!3!}}
⇒\RightarrowNumber of ways to place 3 white flowers in 6 positions=6!3!3! = \dfrac{{6!}}{{3!3!}}
Use the expansion of factorial n!=n(n−1)(n−2)!n! = n(n - 1)(n - 2)!to open the values
⇒\RightarrowNumber of ways to place 3 white flowers in 6 positions=6×5×4×3!3!3! = \dfrac{{6 \times 5 \times 4 \times 3!}}{{3!3!}}
Cancel the same terms from numerator and denominator.
⇒\RightarrowNumber of ways to place 3 white flowers in 6 positions=6×5×43! = \dfrac{{6 \times 5 \times 4}}{{3!}}
Now we multiply this value by number of ways 3 flowers can be arranged i.e. 3!3!
⇒\RightarrowNumber of ways to place 3 white flowers in 6 positions=6×5×43!×3! = \dfrac{{6 \times 5 \times 4}}{{3!}} \times 3!
Cancel the same terms from numerator and denominator.
⇒\RightarrowNumber of ways to place 3 white flowers in 6 positions=120 = 120 … (2)
Number of ways to arrange all the flowers so no white flower is together is given by multiplication of number of ways to arrange red flowers and number of ways to arrange white flowers.
Multiplying equations (1) and (2) we get
Number of ways =120×120 = 120 \times 120

Number of ways =14400 = 14400

Note: Many students skip the step of multiplication with the number of ways in which white flowers can be shuffled within i.e. multiplication with 3!3!. Keep in mind the three white flowers can be put at three different places in 3!3!ways.