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In how many ways can we give 5 apples, 4 mangoes, 3 oranges (fruits of the same species are similar ) to 3 people if each may receive none, one or more.?

In how many ways can we give 5 apples, 4 mangoes, 3 oranges (fruits of the same species are similar ) to 3 people if each may receive none, one or more.?

Grade:12th pass

1 Answers

sohan goswami
23 Points
6 years ago
You have to distribute each type of fruit separately.Let akak, 1≤k≤31≤k≤3, be the number of apples distributed to the kkth person. Thena1+a2+a3=5(1)(1)a1+a2+a3=5The number of ways the apples can be distributed is the number of solutions of equation 1 in the nonnegative integers. A particular solution corresponds to the placement of two addition signs in a row of five ones. For instance,11+1+1111+1+11corresponds to the solution a1=2a1=2, a2=1a2=1, and a3=2a3=2, while+111+11+111+11corresponds to the solution a1=0a1=0, a2=3a2=3, and a3=2a3=2. Therefore, the number of such solutions is the number of ways two addition signs can be inserted into a row of five ones, which is(5+22)=(72)(5+22)=(72)since we must choose which two of the seven symbols (five ones and two addition signs) will be addition signs.The number of ways the mangoes can be distributed is the number of solutions in the nonnegative integers of the equationm1+m2+m3=4(2)(2)m1+m2+m3=4where mkmk, 1≤k≤31≤k≤3, is the number of mangoes distributed to the kkth person. The number of such solutions is$$\binom{4 + 2}{2} = \binom{6}{2}$$The number of ways the oranges can be distributed is the number of solutions in the nonnegative integers of the equationo1+o2+o3=3(3)(3)o1+o2+o3=3where okok, 1≤k≤31≤k≤3, is the number of oranges distributed to the kkth person. The number of such solutions is$$\binom{3 + 2}{2} = \binom{5}{2}$$To find the number of ways of distributing the three types of fruit, multiply the above results.

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