A shoping mall is running a scheme: Each packet of detergent SURF contains a coupon which bears letter of the word SURF, if a person buys at least four packets of detergent SURF, and produce all the letters of the word SURF, then he gets one free packet of detergent.If a person buys 8 such packets at a time, then the number of different combinations of coupon he has isIf person buys 8 such packets, then the probability that he gets exactly one free packets isIf a person buys 8 such packets, then the probability that he gets two free packets is

Let in 8 coupons S, U, R, F appears x1, x2, x3, x4 times. Then x1+x2+x3+x4=8, where x1,x2,x3,x4≥0.We have to find non-negative integral solutions of the equation. The total number of such solutions is 8+4−1C4−1=11C3=165.If a person gets at least one free packet, then he must get each coupon at least once, which is equal to number of positive integral solutions of the equation. The number of such solutions is  8−1C4−1=7C3=35. Then, the probability that he gets exactly one free packet is (35 - 1)/165 = 34/165The probability that he gets two free packets is 1/11C3=1/165.Let in 8 coupons S, U, R, F appears x1, x2, x3, x4 times. Then x1+x2+x3+x4=8, where x1,x2,x3,x4≥0.We have to find non-negative integral solutions of the equation. The total number of such solutions is 8+4−1C4−1=11C3=165.If a person gets at least one free packet, then he must get each coupon at least once, which is equal to number of positive integral solutions of the equation. The number of such solutions is  8−1C4−1=7C3=35. Then, the probability that he gets exactly one free packet is (35 - 1)/165 = 34/165The probability that he gets two free packets is 1/11C3=1/165.Let in 8 coupons S, U, R, F appears x1, x2, x3, x4 times. Then x1+x2+x3+x4=8, where x1,x2,x3,x4≥0.We have to find non-negative integral solutions of the equation. The total number of such solutions is 8+4−1C4−1=11C3=165.If a person gets at least one free packet, then he must get each coupon at least once, which is equal to number of positive integral solutions of the equation. The number of such solutions is  8−1C4−1=7C3=35. Then, the probability that he gets exactly one free packet is (35 - 1)/165 = 34/165The probability that he gets two free packets is 1/11C3=1/165.