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This is one of our many quizzes on Permutations. Click on this to access more such competitive quizzes on Permutations and Combinations.

Permutations Quiz 2

mathematics

 In class, there are 10 girls. All girls are of different heights. In how many ways can a team of 3 girls be selected out of these 10 girls, in which the first girl is always shorter in height than the second girl and the second girl is shorter than the third girl and so on?

 Use the combination formula, $^{n}{{C}_{r}}=\dfrac{n!}{r!(n-r)!}$, here the total number of girls is $n=10$, 3 girls has to be selected, so $r=3$.

$\begin{align} &nbsp;&nbsp;& 3\text{ girls has to be selected from 10, so the required number of ways:} \\ &nbsp;& ^{10}{{C}_{3}}=\dfrac{10!}{3!(10-3)!} \\ &nbsp;& {{\Rightarrow }^{10}}{{C}_{3}}=\dfrac{10\times 9\times 8}{6} \\ &nbsp;& {{\therefore }^{10}}{{C}_{3}}=120 \\ \end{align}$

 Blocks of letters are used to form different words. How many different three-letter words can be formed by using the blocks of letters from the word MATHEMATICS?

 Three letter words can be formed either from choosing all three different letters or by choosing two same and one different letter, add these two cases for the total number of ways.

\[\begin{align} &nbsp;&nbsp;& \text{Three letter words can be formed in the following manner:} \\ &nbsp;& \left( \text{i} \right)\text{ All three different letters:} \\ &nbsp;& \text{All three different letters can be chosen in}{{\text{ }}^{8}}{{C}_{3}}\text{ } \\ &nbsp;& =\text{ }56\text{ ways and can be arranged in 3! ways}\text{.} \\ &nbsp;& \text{Therefore the total number of words formed }=56\times 3!=336\text{ }\!\!~\!\!\text{ } \\ &nbsp;& \left( \text{ii} \right)\text{ Two same, one different:} \\ &nbsp;& \text{Letters M, A, T repeat once in MATHEMATICS}\text{. } \\ &nbsp;& \text{The letter that repeats can be selected in 3 ways}\text{.} \\ &nbsp;& \text{The remaining letter can be selected in 7 ways}\text{.} \\ &nbsp;& \text{And these three letters can be arranged in 3 ways}\text{.} \\ &nbsp;& \text{Total number of words formed = 3 }\!\!\times\!\!\text{ 7 }\!\!\times\!\!\text{ 3 = 63 ways}\text{.} \\ &nbsp;& \text{From both the cases we get 3}36+63=399\text{ ways }\!\!~\!\!\text{ } \\ \end{align}\]

 If the different permutations of the word PARAKEET are listed as in the dictionary, then what will be the rank of the word PARAKEET?

 First write the given word in alphabetical order then find out the number of words that start with a superior letter and then solve the same problem, without considering the first letter.

\[\begin{align} &nbsp;&nbsp;& \text{The letters of the word PARAKEETin the alphabetical order are A, A, E, E, K, P, R, T}\text{.} \\ &nbsp;& \text{Now, number of words beginning with the letter A }=\dfrac{7!}{2!} \\ &nbsp;& \text{Number of words beginning with the letter E }=\dfrac{7!}{2!} \\ &nbsp;& \text{Number of words beginning with the letter K }=\dfrac{7!}{2!\times 2!} \\ &nbsp;& \text{Number of words beginning with the letters PAA }=\dfrac{5!}{2!} \\ &nbsp;& \text{Number of words beginning with the letters PAE }=5! \\ &nbsp;& \text{Number of words beginning with the letters PAK }=\dfrac{5!}{2!} \\ &nbsp;& \text{Number of words beginning with the letters PARAE = }3! \\ &nbsp;& \text{The next word that comes is PARAKEET} \\ &nbsp;& \text{So, the rank of the word PARAKEET = }\dfrac{7!}{2!}+\dfrac{7!}{2!}+\left[ \dfrac{7!}{2!\times 2!} \right]+\dfrac{5!}{2!}+5!+\dfrac{5!}{2!}+3!+1 \\ &nbsp;& =6300+240+6+1=6547 \\ \end{align}\]

 What is the number of permutations of alphabets J, U, L, I, E, T, M, Y, R, A, H, O, B such that none of the words JULIET, MYRA, HOB occur as consecutive letters?

 The number of permutations will be by subtracting the number of arrangements formed by taking the three words, JULIET, MYRA, HOB separately, taking two of them together, and taking all three together.

\[\begin{align} &nbsp;&nbsp;& \text{The required number can be calculated by}~ \\ &nbsp;& \text{Total arrangements - one of them formed }\left( \text{Juliet + Myra + Hob} \right)\text{ } \\ &nbsp;& \text{+ pair formed}\left( \text{Juliet Myra + Myra Hob+Juliet Hob} \right)\text{ - all 3 formed }\left( \text{Juliet Myra Hob} \right) \\ &nbsp;& =13!-\left( 8!+9!+10! \right)+\left( 5!+7!+6! \right)-3! \\ &nbsp;& ~=6227020800-\left( 4032000 \right)+\left( 5880 \right)-6=6222994674 \\ \end{align}\]

 Consider the number 5789. Form all the possible permutations of this number by changing the order of its digits in it. Add these permutations to get a number A. How many multiples of 1111 less than 10000 divide A?

 Find the sum of four random digits then for five digits by the formula \[n!\times 111..(\text{n times})\times \left( a+b+c+...\text{n terms} \right)\] for the number then add all possible permutations to get $''A''$.

\[\begin{align} &nbsp;&nbsp;& \text{The sum of all permutations of a four digit number }abcd\text{ is given by } \\ &nbsp;& 3!\times 1111\times \left( a+b+c+d \right) \\ &nbsp;& \text{Similarly, for a five digit number }abcde\text{ it is equal to }4!\times 11111\times \left( a+b+c+d+e \right). \\ &nbsp;& \text{Sum of all the permutations of 5}789 \\ &nbsp;& =3!\times 1111\times \left( 5+7+8+9 \right)=3!\times 1111\times 29=193314 \\ &nbsp;& \text{Therefore }A=193314 \\ &nbsp;& \text{Therefore the multiples of }1111\text{ that will divide }A\text{ are } \\ &nbsp;& 1111,\text{ }2222,\text{ }3333\text{ and }6666 \\ &nbsp;& \text{There are 4 multiples of 1111 less than 10000 that divide }A. \\ \end{align}\]