Question

If all the letters of the word "PATLIPUTRA" are arranged then find the number of words which can be formed when:

(1.) Relative order of vowels remain same
(2.) Vowels occupy even places

Class 11 P and C
Please solve

Answer 1

Answer:

Step-by-step explanation:

1) For fixed relative order of vowels we need to select 4 spaces without rearranging(AIUA) and then rearrange the remaining 6 letters(PTLPTR) in 6 spaces.

Note: To rearrange n objects where object a is repeated p times and b is repeated q times and so on = n!/(p! * q! * ..)

Thus,

Number of words

= ¹⁰C₄ + [⁶c₆ * 6!/(2!*2!)]      

= 210 + 180

= 390

2) Total number of letters = 10

Total number of vowels = 4 (AIUA)

Number of possible even places = 5

Here, we select 4 spaces(AIUA) out of 5 even spaces and rearrange the vowels. After that we rearrange the 6 consonants(PTLPTR) in remaining spaces.

Note: To rearrange n objects where object a is repeated p times and b is repeated q times and so on = n!/(p! * q! * ..)

Thus,

Number of words

= [⁵C₄ * 4!/2!] + [⁶C₆ * 6!/(2! * 2!)]

= 60 + 180

= 240

Answer 2

$\large\underline{\sf{Solution-1}}$

The given word is PATLIPUTRA

Now, there are 4 vowels [A, A, U, I] in the word PATLIPUTRA.

So, number of ways in which 4 vowels can arrange themselves on their respective places is $\rm \: \dfrac{^4P_4}{2!}$ ways.

And number of ways in which remaining 6 consonants [P, P, T, T, R, L] on their respective places is $\rm \: \dfrac{^6P_6}{2! 2!}$ ways.

So, total number of ways relative order of vowels remain same is

$\rm \:= \dfrac{^4P_4}{2!} \times \dfrac{^6P_6}{2!2!}$

$\rm \: = \dfrac{24}{2} \times \dfrac{720}{4}$

$\bf\implies \:Number\:of\:ways\:=\:2160$

$\large\underline{\sf{Solution-2}}$

Now, there are 5 even places and 5 odd places. So, 4 vowels [A, A, U, I] can be arranged themselves in 5 places in $\dfrac{^5P_4}{2!}$

Now, remaining 6 consonants [P, P, T, T R, L] can arranged theselves in $\rm \: \dfrac{^6P_6}{2!2!}$ ways.

So, total number of ways vowels occupy even places is

$\rm \:= \dfrac{^5P_4}{2!} \times \dfrac{^6P_6}{2!2!}$

$\rm \: = \dfrac{120}{2} \times \dfrac{720}{4}$

$\bf\implies \:Number\:of\:ways\:=\:10800$