9 (C) (a) You give your students the following problem: "You just bought 30 beads of which 18are gold and identical and 12 are silver and identical. You want to line them up in arow. How many different arrangements of the beads can you make?" Elizabethsolves the problem by finding the number of distinguishable permutations of 30objects, 18 alike of one kind and 12 alike of another kind. That is, she uses theresult of the previous problem. Natalie solves the problem by finding the number ofcombinations of 30 different objects taken 18 at a time. Who is correct?(b) Does the number of permutations of n objects, ralike of one kind and n r alike of anotherkind, always equal the combinations of n different objects taken r at a time? Explain.

Given information: Total number of beads = 30 Number of golden beads = 18 Number of silver beads = 129) C) a) Determine the different arrangements of beads using Elizabeth's method: Total number of different arrangements=P3030P1818×P1212=30!30-30!18!18-18!×12!12-12!=30!12!×18!=86,493,225 ways Determine the different arrangements of beads using Natalie's method: Total number of different arrangements=C1830=30!18!×30-18!=30!18!×12!=86,493,225 ways Number of different arrangements obtained from both the methods are same, hence both of them are correct.b) The formula for permutations of n objects, r alike of one kind and n - r alike of another kind is given as: Permutation of n objects r alike of one kind and n-r alike of another kind =n!n1!×n2!=n!r!×n-r! Combination of n different objects taken r at a time =Crn=n!r!×n-r! The formula for both the methods can be simplified to same formula, hence the number of permutations of n objects, r alike of one kind and n - r alike of another kind, always equal the combinations of n different objects taken r at a time. It is applied only till the number of different type of objects is 2, if there are more than two kinds of objects than this formula will not be valid.